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ETHEidgen6ssische Technische HochschuleZurichInstitut fur Informatik
Niklaus WirthA COLLECTIONOF PASCAL PROGRAMS
July 1979- 33
E T H
EIDGENOSSISCHE TECHNISCHE HOCHSCHULE
ZORICH
INSTITUT FUR INFORMATIK
NIKLAUS WIRTH
ACOLLECTION OF PASCAL PROGRAMS
JULY 1979 33
Address of the author:
Institut fUr InformatikETH-ZentrumCH-8092 ZUrich
® 1979 Institut fUr Informatik der ETH ZUrich
Conlents
Contents
O. Preface
1. Integer arithmetic
power, divide, gcd-Icm, binary ged, j-sqrt
2. Integer arithmetic and arrays
palindromes, magic squares, powers of two, fmctions,harmonic function, prime numbers, sieve of Eratosthenes
3. Real (floating-point) arithmetic
sum10000, complex multiplication, Fibonacci numbers
4. Analytic functions and iteration
sqrt, logarithm, reciproc<ll value,exp, sin, cos, arcsin, arctan, In
6. Text processing
printerplot, edit, count wordiengllls, crunch, hilparade
6. Recursion
permute, infix-poslfix, Hilbert curves, Sierpinsl~i curves
7. Sorting arrays
slraioht insertion, binary insertion, Shellsort,straiol1t selection, l1eapsort, bubblesort, shal~ersort,
quicksort, rnergesort
8. Sequential sorting
nalural merge, balanced merge, polyphase merge
9. "Problem solving", backtracl~ing
eight queens, nonrereating soqUGllces, sLIm of cuhes,knight's tour, st8ble marriages, optimal selection
10. List and tree structures, pointers
search and insertion in ordered list,search and insertion in reordering Iisl,topological sorting, insertion and deletion in binnry troe,insertion and deletion in bnl,lIlced tree,insertion and deletion in G-lree, optimal senrcll tree
"11. Cross-reference generators
cross-reference gp.ner8.tor lIsing binary treecross-reference generalor using hasll table
12. Syntax an<:tlysis
Syntax analyser and scanner for 1,,1ngu8ge PLIO
2
o. Preface
Pascnl prn'jrnrns
This is a collecHon of a wide variety of PaSC81 programs. They ranne in complexityfrom simple examples used in introductory courses to illustrate desiqn principles andInnguage features to intricole eX8111ples discussecl in courncs on 81~Jorithrns ond dntnstructures. The programs, however, are WOllped nccorcling to subjecl rnaller millerlhon complexity. Many me tahen from the literature listed below, where they areexplained and analyzed in detail.
The main purpose of Ihis bool<let is to provide the tEitlcher of proDrnmming willl [\condensed collection of exemplary progrmns ancl thereby to exhibit a preferred slyleof programming using 8 structured Iflngll<1~le. AI the same lime, the bool,let may servo(IS a guide in inventing other, perhaps similar exercises. Lnstly, it may be a helpfulreference to some widely used, fundamental algorilhlflS, formulated in detail in awidely avniloble language.
neferences
N. Wirth, Systematic prowamming. Prell!ice+lall, Inc. 1973.
I\/oorilhms -I- c/vta structures::: programs. Prentice-Hall, Inc. 1975.
1. Integer arithmetic
Integer arithmetic 3
1. Raise integer to a positive power. Repeot reading pairs of inlegers, until youencounter a O. Indicate invariant of loop.
PROGRAM power(input, output);VAR 8, b: integer;
FUNCTION power (x,n: integer): integer;VAR w,z,i: integer;
BEGIN w := x; i := n; z := 1;WHILE i # 0 DOBEGIN (. z+w'ti = x'tn +)
IF odd(i) THEN z := z'w;w := sqr(w); i := i DIV 2
END;power := Z
END (+ power +) ;
BEGIN read(a);WHILE a # 0 00BEGIN read(b); writeln(a, b, power(a,b»);
read(a)END
END.
2. Divide an integer by a natural number, using operations of addition, subtraction,doubling and halving only. Repeal reading pairs of inlegers, until you encounter a O.For each pair, print dividend, divisor, quotient, and remainder. Indicate invariant ofloop.
PROGRAM divide(input, output);VAR a,b,Cf,r: integer;
PROCEDURE divide (x,y: integer; VAR z,a: integer);VAR q,r,w: inleger;
BEGIN r := x; w := y; q := 0;WHILE w <= r DO w := 2'w;WHILE w # y DOBEGIN (+ x = q'w + r t) w:= w DIV 2; q := 2'q;
IF w <= r THENBEGI!'I r := r-w; q := q+ 1END
E!'ID;Z := q; a := r
END ('divide t) ;
BEGIN read(a);WHILE a it 0 00BEGIN rend(b); divide(a,b,q,r); writeln(a, b, q, r);
read(a)END
END.
4 Pascal programs
3. Compute the greatest common divisor (gcd) and the lowest cornman mUltiple (Icm)01 two natural numbers by using addition and subtraction only. Note thaI ged(rn,n) °lcm(m,n) =m"n. Repeat reading pairs 01 Integers, until you encounter a O. For eachpair, print the arguments, the gcd and lhe tern. Indicate the loop invariant.
PROGRAM gcdlem(input, oUlput);VAR a,b,c,d: integer;
PROCEDURE gcd(x,Y: integer; VAR u,v: integer);VAR a,b,c,d: integer;
BEGIN a := x; c := x; b := y; d := y;WHILE a # b DOBEGIN (Ogcd(a,b) = gcd(x,y) AND aOd + b"c =2'x"y')
IF a ) b THENBEGIN a := a-b; c := c+dEND
ELSEBEGIN b := b-a; d := d+cEND
END;u := a; v := (c+d) DIV 2
END ("gcdmult") ;
BEGIN read(a);WHILE a # 0 DOBEGIN read(b); gcd(a,b,c,d); writeln(a, b, c, d);
read(a)END
END.
4. Compute the greatest common divisor (ged) 01 two natural numbers. Use addition,subtraction, doubling and halving only.
PROGRAM binarygcd(output);VAR a.b: integer;
FUNCTION gcd (x,y: integer): integer;VAR u,v,d, [I,b,I<: integer;
BEGIN u := x; v := y; a := 0; b := 0;WHILE NOT odd(u) DO
BEGIN u := u DIV 2; a := a+1END;
WHILE NOT odd(v) DOBEGIN v := v DIV 2; b := b+1END; .
IF a<b THEN k := a ELSE k := b;d := u - v;WHILE d # 0 DOBEGIN REPEAT d := d DIV 2 UNTIL odd(d);
IF d<O THEN v := -d ELSf; u := d;d := u - v
END;WHILE k>O DOBEGIN u := 2·u; k := k-1END;ged := u
END;
Integer arithmetic
BEGIN read(a);WHILE a # 0 DOBEGIN read(b); writeln(a, b, gcd(a,b»; read(a)END
END.
5
5. Compute the largest integer less or equal to the square rool of a given integer(due to Hoare).
PROGRAM isqrt(inpul,output);VAR n,a2,b2,ab,t: integer;
BEGIN read(n);WHILE n )= 0 DOBEGIN a2 := 0; ab := 0; b2 := 1; writeln(" n =", n);
WHILE b2 <= n 00 b2 := 4 ·b2;WHILE b2 # 1 DOBEGIN (. a2+2·ab+b2 > n, 0 <= a2 <= n, sqr(nb) =a2·b2 .)
ab := ab DIV 2; b2 := b2 DIV 4; t := a2 + 2t ab .,. b2;IF t <= n THEN
BEGIN a2 := t; ab := ab + b2;END
END;writeln(a2,ab,b2); read(n)
ENDEND.
6 Pascal programs
2. Integer arithmetic and arrays
1, Find all integers between 1 and 1000 whose squares are palindromes. Examples:11t2 = 121, 22t2 = 484.
PROGRAM palindromes(oulpul);VAR i,i,l,n,r,s: integer;
p: boolean;d: ARRAY [1 .. 10] OF integer;
BEGIN n := 0;REPEAT n := n+1; s := n·n; I := 0;
REPEAT I := 1+1; r := S DIV 10;d[l] := s - 10·r; s := r
UNTIL s :: 0;i '- l' j '- I'REPEAT'~ ;= d[i]=d[j];
i := i+1; j := j-1UNTIL (i> =j) OR NOT p;IF P THEN wrileln(n,n·n)
UNTIL n = 1000END.
2. Compute and print magic squares of order 3, 6, 7, ...
PROGRAM magicsquare (output);CONST lim = 11;VAR i,j,x,nx,nsq,n: integer;
m: ARRAY [1 ..lim, 1..lim] OF integer;
PROCEDURE gelsquare;BEGIN x ::: 0; nsq := sqr(n);
i := (n+1) DIV 2; j := n+1;REPEAT nx := x + n; j := j-1;
x := X+ 1; m[i,j] := x;REPEAT i :'" i+1; IF i >n THEN i := 1;
j := j+ 1; IF j > n THEN j := 1;x := x+ 1; m[i,j] := x
UNTIL x = nxUNTIL x :: nsq
END (tgetsquare·) ;
PROCEDURE prinlsquare;BEGIN
FOR i := 1 TO n DO .BEGIN FOR j := I TO n DO wrile(m[i,j]: 6);
wrilelnEND;wrileln
END (tprinlsquare·) ;
BEGIN n := 3;REPEAT getsquare; prinlsquare; n := n+2UNTIL n > lim
END.
Integer arithmetic and arrays 7
3. Compute a table of positive and negative powers of 2. Exponents range from 1 10,say, 64. Do not truncate any digilsl
PROGRAM powersoftwo(output);CONST m =31; n = 100; (. m = n'log(2) ')VAR exp,i,j,l: integer;
c,r,l: integer;d: ARRAY [O..m] OF inleger; ('positive powers')f: ARRAY [l ..n] OF inleger; ('negative powers')
BEGIN I := 0; r := 1; d[O] := 1;FOR exp := 1 TO n DOBEGIN (. compute and print 2' 'exp') c:= 0;
FOR i := 0 TO 1 DOBEGIN t := 2'd[i] + c;
IF I )= 10 THENBEGIN d[i] := t-10; c := 1END
ELSEBEGIN d[i] := I; c := 0END
END;IF c > 0 THEN
BEGIN I := 1+ 1; d[l] := 1END'
FOR i :'= m DOWNTO I DO write(" ");FOR i := I DOWNTO 0 DO write(d[i]:1);write(exp:S," .");('compute and print 2"(-exp) .)FOR j := 1 TO exp-1 DOBEGIN r := 10'r + I[j];
r[n := r DIV 2; r ::: r - 2'f[j]; wrile(f[j):l)END'r[exp'] := 5; wrileln("S"); r := 0
ENDEND.
8 Pascal programs
4. Compute a table of exact fractions 1/2, 1/3, ... , 1/64. If the fraction has a period,print an apostrophe in front of its first digit and truncate after its last digit.
PROGRAM fractions(output);(" fractions to the base b ")
CaNST b = 10; max =64;VAR i,',n,q,r: integer;
a. f: ARRAY [O..max] OF integer;BEGIN FOR n := 2 TO max DO
BEGIN I := 0; r := 1;FOR i := 0 TO n-1 DO a[i] := 0;REPEAT I := 1+1; 8[r] := I;
r := b"r; q := r DIV n; r := r - q"n; f[l] := q;UNTIL a[r] # 0;wrile(n, .. ", ".");FOR i := 1 TO a[r]-1 DO write(f[i]:1);IF a[r] > 1 THEN write(" ...... );FOR i := a[r] TO I DO write(f[i]:1);writeln
ENDEND.
5. Compute the harmonic function H(n) 1 + 112 + 113 + ... + 1In with m digitsaccuracy.
PROGRAM harmonic(input,output);CaNST lim = 100;VAR i,k,m,n,c,r,q,sum: integer;
d,s: ARRAY [O..lim] OF integer;BEGIN read(m,n);
IF (01)0) AND (m(lim) THENBEGIN d[O] := 0; s[O] := 1;
FOR i := 1 TO m DO sri] := 0;FOR k := 2 TO n DOBEGIN ("compute 11k)" r := 1;
FOR i := 1 TO 01 DOBEGIN r := 10"r; q := r DIV k; r := r-q"k; d[i] := qEND ; 0
IF (i0"r DIV k) >= 5 THEN d[m] := d[m]+1; ("round)·write(" 0."); ("intermediate output") ,FOR i :=1 TOm DO write(d[i]:1);writeln;("compule s := s + i/k)" c:= 0;FOR i := m DOWNTO 0 DO
BEGIN sum := s[i]+d[i]+c;IF sum >= 10 THEN
BEGIN sum := sum-10; c :=END
ELSE c := 0;sri] := sum
ENDEND;write(" ", s[0]:1, .....);FOR i := 1 TO m DO write(s[i]:1);writeln
ENDEND.
Integer arithmetic and arrays
6. Compute a table of the first n prime numbers. Print m numbers per line.
PROGRAM primes(output);CONST n = 1000; n1 = 33; m = 20; (*n1 - sqrt(n)*)VAR i,k,x,inc,lirn,square,l: integer;
prim: boolean;p,V: ARRAY [1..n1] OF integer;
BEGIN I := 0;x := 1; inc := 4; lim := 1; square := 9;FOR i := 3 TO n DOBEGIN (*find next prime*)
REPEAT x := x+inc; inc := 6-inc;IF square <= x THEN
BEGIN lim := lim+ 1;v[lim] := square; square := sqr(p[lim+1])
END;k := 2; prim := true;WHILE prim AND (k<lim) DOBEGIN k := k+1;
IF v[k] < x THEN v[k] := v[k] + 2*p[k];prim := x # v[k]
ENDUNTIL prim;IF i <= n1 THEN p[i] := x;wrile(x:6); I := 1+1;IF I = m THEN
BEGIN wrileln; I := 0END
END;writeln
END.
9
10 Pascal programs
7. Compute a table of the first n prime numbers. Print m numbers per line. Use themethod of the sieve of Eratosthenes.
PROGRAM primes(output);CONST m = 100; n = 10000; m = 20; h = 58;VAR x, inc, i,k, x1,x2, lim, square, a,b,l: integer;
P,V: ARRAY [Lm] OF integer;sieve: SET OF O..h;
BEGIN I := 0;x := 1; inc := 4; lim := 1; square := 9;xi := 0; x2 := 0; sieve := [O..h];p[1]:= 2; p[2]:= 3;FOR i := 3 TO n DOBEGIN ('"find next prime)'"
REPEAT x := x+inc; inc := 6-inc;IF x >= square THENBEGIN lim := Iim+ 1; a := square; b := 2*p[lim];
WHILE a <x2 DOBEGIN sieve := sieve - [a-x1]; a := a+bEND'
v[lim] ;= a; square := sqr(p[lim+ 1])END;IF x >= x2 THENBEGIN ('"construct new sieve)·
xi := x2; x2 := x2+h; sieve := [O..h];FOR k := 3 TO lim DOBEGIN a := v[k]; b := 2·p[k];
WHILE a < x2 DOBEGIN sieve := sieve - [a-xi]; a := a+bEND;
v[k] := aEND
ENDUNTIL x-xi IN. sieve;IF i <= m THEN p[i] := x;write(x:6); I := 1+1;IF 1= m THEN
BEGIN writeln; I := 0END
END;writeln
END.
Real (floating-point) arilhmelic
3. Real (floating point) arithmetic
1. Compute the sum 1 - 1/2 + 1/3 - 1/4 + ... - 1/10000 in four different ways:1. proceed strictly from lerr to right,2. sum positive and negative terms separately,3. proceed strictly from right to left,4. as in 2., but from right to lefl.
Explain the differences in the results.
PROGRAM sum10000(output);CONST n = 10000;VAR i: integer; x, y, s1, s2p, s2n: real;
BEGIN i := 1;sf := 0; s2p := 0; s2n := 0;REPEAT x := 1.0/i; y := 1.0/(i+1);
s1 := s 1 + x - y;s2p := s2p + x; s2n := s2n + y;'i := i+2
UNTIL i > n;write (sl, s2p-s2n);i := n;s1 := 0; s2p := 0; s2n := 0;REPEAT x := 1.0/(i-1); Y := 1.0/i;
s 1 := s 1 + x - y;s2p := s2p + x; s2n := s2n + y;i := ;-2
UNTIL i = 0;writeln(s 1, s2p-s2n)
END.
11
2. Multiply the complex number z = 5/13 + 12/13i 50 limes with the complexnumber w = (0.6 + 0.8i). Print intermediate products and tile square of theIr absolutevalue. Note thai Izi = Iwl = 1.
PROGRAM complexmulI(output);CONST u = 0.6: v =0,8;VAR i,j: integer; x,x1,Y: real;
BEGIN x := 5/13; y := 12/13;FOR i ::: 1 TO 50 DOBEGIN FOR j := 1 TO 10 DO
BEGIN (. (x+iy) := (x+iy) • (u+iv) ')x'I := x'u - y·v; y := y·u + x'v; x := xl
END;wrileln( x,y ,sqr(x)·t-sqr( y»
ENDEND.
12 Pascal programs
3. Compute the Fibonacci numbers F( 1) ... F(N) in two different ways:1. By repealed addition according to F(n) = F(n-1) + F(n-2), F(O) = F(1) 1,2. Using the formula F(n) ~ (phitn)/sqrt(5), where phi = (1+sqrt(5))/2.
Terminale as soon as the two results differ.
PROGRAM fibonacci(oulpul);CONST rool5 = 2.236068;VAR it fibO, fib1, fib3, t: integer;
phi,lib2: real;BEGIN phi := (1.0+root5)/2;
i := 0; fibO := 1; fib 1 := 0; fib2 := 1.0 / root5;REPEAT i := i+1;
I := fibO+fib I; fibO := fib1; fib1 := t;fib2 ;= fib2 • pili; fib3 := lrunc(fib2 + 0.5);wrHeln(i, fib1, fib3)
UNTIL fib1 # fib3END.
Analytic functions and iteralion
4. Analytic functions and iteration
CONST eps = 1E-8;
FUNCTION sqrt (x:real): real;VAR a,e: real; (+ 0<x<2 +)
BEGIN Cl := x; e := 1.0-x; (+ abs(c)< 1 +)REPEAT (+ at2 = bO'(1-c) =) (a+(1+e/2»t2 = b+(i-e)+(1+e/2)t2 =
b+( 1-0.7S+(et2) - 0.25+(ct3» +)a := a+( 1.0 + O.S·c);
(O' at2 ::: b+(1-0.7S+(ct2) - 0.25+(c t 3» =::: b+(1-(ct2)·(O.75 + 0.2S+c) +)
e := sqr(c) + (0.75 + 0.2S+c);(+ at2 = b+(1-c) .)
UNTIL abs(c) < eps;sqrl := a
END;
FUNCTION log (x: real): real;VAR a,b,s: real; (0 1<=x<2 0)
BEGIN a := x; b :::: 1.0; s := 0;REPEAT (+ log(x) = s + b+log(a), b<=1, 1<=a<2 +)
a := sqr(a); (0 log(x) = s + b+log(sqrl(a», 1 <=a<4 +)b := O.S+b; (+ log(x) = s + b'log(a) +)IF a >::: 2.0 THENBEGIN (+ 2<a<4 +) s := s+b (+ log(x) = s + (1-b)+log(a) ');
a :::: O.S'aEND
UNTIL abs(b) < eps;log :::: s
END;
FUNCTION recip (x: reat): real;VAR a,e: real; (+ 0<x<2 +)
BEGIN a := 1.0; e := 1.0 - x;REPEAT (+ a*x ::: i-c, abs(c)< 1 +)
a := a·(1.0+e); (+ x*a = (1-c)+(1+e) = 1 - ct2 +)c := sqr(c); (+ x+a ::: 1-c +)
UNTIL abs(c) < eps;recip := a (* recip = 1/x +)
END;
13
14 Pascal programs
Compute analytic functions as truncated sums. Determine the recurrence relations oftheir terms.
PROGRAM recurrence(input,outpul);VAR i,n: inleger; x,Y,s,t: real;
BEGINwriteln(" exp"); n := 5;REPEAT read(x); y := 1.0; i := 0; t := 1.0;
REPEAT i:= i+1; t:= t·x/i;y := y + t
UNTIL y+t = y;writeln(x,Y,i); n:= n-1
UNTIL n = 0;
writeln(" sin"); n := 5;REPEAT read(x); y := x; i := 1; s := sqr(x); t := x;
REPEAT i := i+2; t := -t's/((i-1 Vi);y := y + I
UNTIL y+t = y;wrileln(x,Y, i DIV 2); n:= n-1
UNTIL n = 0;
writeln(" cos'!); n := 5;REPEAT read(x); Y := 1.0;i := 0; s := sqr(x); 1:= 1.0;
REPEAT i := i+2; t := -t's/((i-1 )'i);y := y + t .
UNTIL y+t = y;wrileln(x,y, i DIV 2); n:= n-1
UNTIL n = 0;
wrileln(" arcsin"); n := 5;REPEAT read(x); y := x; i := 1; s := sqr(x); I := x;
REPEAT i := i+2; t := t"s"sqr(i-2)/((i-1)"i);y := y + t
UNTIL y+t = y;writeln(x,Y, i DIV 2); n:= n-1
UNTIL n = 0;
wrileln(" arctan"); n := 5;REPEAT read(x); y := x; i := 1; s := sqr(x); t := x;
REPEAT i := i+2; t := -1's'(i-2)/i;y := y + I
UNTIL y+l = y;wrileln(x,Y. i DIV 2); n:= n-1
UNTIL n = 0;
writeln(" In"); n := 5;REPEAT read{x); x := x-1.0; y := x; t := x; i := 1;
REPEAT i := i+1; t := -t'x"(i-1)/i;y := y + t
UNTIL y+t = y;wrileln(x+ 1.0, Y, i); n:= n-1
UNTIL n = 0;END.
5. Text processing
Texl processing 15
1. Plot the function f(x) = exp(-x) • cos(2·pj·x) with your line printer in the range x= 0 ... 4. Use 32 lines for the unit coordinate.
PROGRAM printerplot(output);CONST xscale = 32;
yscale = 50; ysl1Ht = 65;twopi = 6.2831833071796;
VAR i,k,n: integer;x,y: real;
BEGIN n := 0; (. n = x position .)REPEAT x := n / xscale;
y := exp(-x)·cos(x·lwopi); k := round(y·yscFlle);i := 0; write(" "); (. i = no of chars in fine .)IF k < 0 THENBEGIN write(" ": yshifl+k); write("·");
k := -k-1; IF k > 0 THEN write(" ":k);write("I")
END ELSEBEGIN write(" ": yshifl);
IF k > 0 THENBEGIN write(',!,'); k := k-1;
IF k > 0 THEN write(" ":k)END;write("·")
END;writeln; n := n+1
UNTIL n >96END.
2. Read a text and cOLint the number of words with length 1, 2, ... , 20, and those withlenglh greater than 20. Words are separated by blanks or ends of lines.
PROGRAM wordlenglhs(input,output);VAR i,k: inleger;
eh: char;coun!: ARRAY [1 ..21] OF integer;
BEGINFOR i := 1 TO 21 DO counl[i] := 0;WHILE NOT eof(input) DOBEGIN read(ch);
IF ("a"<=eh) AND (eh<="z") THENBEGIN (·new word)· k := 0;
REPEAT k ::: k+1; read(ch)UNTIL (ell("a") OR ("z"<ch) ;IF k > 20 THEN 1< := 21;eOllnt[k] := counl[k] + 1
ENDEND;wrileln;wrileln(" length count");FOR i:= TO 21 DO writsln(i,count[i])
END.
16 Pascal programs
3. Read a text and produce a copy with flushed left and right margins. Place a fixednumber of characters (say, length = 72) in each line, and distribute blanks as word
• separators accordingly.
PROGRAM edil(inpul,oulput);CONST length = 72;VAR ch: char;
i,m,k,lim: integer;line: ARRAY [1..136] OF char;index: ARRAY [1 .. 68] OF integer;
PROCEDURE sellina;VAR i,j,h,s: integer;
spaces, q,I,r: integer;BEGIN IF m=O THEN
BEGIN (·word is longer than line)· m := 1; index[m] := limEND;
j := 0; write(" "); ('printer control)'IF m ) 1 THENBEGIN spaces := lim - index[m];
q := spaces DIV (m-1); r := spaces - (m-1)'q;I := (m-r) DIV 2; r := l+r; ('distribute spaces)·i '- o·REPEAT i := i+1; s := index[i];
REPEAT j := j+ 1; wrile(line[iJ}UNTil j = s;FOR h := 1 TO q DO wrile(" or);IF (I<=i) AND (i<r) THEN wrile(" or);
UNTIL i =m-1END;s := index[m] -1;REPEAT j := j+1; write(line[j])UNTIL j = s;j := 0; wrileln;FOR h := index[m]+1 TO lim DO
BEGIN j := j+ 1; line[j] := line[h]END;
k := j; m := 0END (·selline)· ;
BEGIN lim := lenglh+1;k := 0; (·I~ = no. OF characters IN line)'m := 0; (·m = no. OF complete words IN line)·WHILE NOT eof(input) DOBEGIN read(ch);
IF ch # " " THENBEGIN (·next word)'
REPEAT k := kt1; Hne[k] := ch; read(ch);IF k = lim THEN setline
UNTIL ch = " or;k := k+l; line[k] := " or;m:= m+1; index[m]:= k;IF k = lim THEN salline
ENDEND;write(" "};FOR i := 1 TO I~ DO write(line[i]);writeln
END.
Text processing 17
4. Read a text and replace any sequence of one or more blanks by a single blank.
PROGRAM crunch(input,oulpul};CONST blank = .. It;VAR ch: char;
BEGINWHILE NOT eof(input} DOBEGIN read(ch); write(blanl~}; ('printer control')
WHILE ch = blank DO read( ch};WHILE NOT eoln(input) DOBEGIN
REPEAT write(ch}; read(ch}UNTIL ch =blank;wrile(blank};WHILE (ch:blank) AND NOT eoln(input} DO read(ch)
END;writeln; read(ch)
ENDEND.
18 Pascal . programs
5. A record company conducts a poll 10 evaluale its products. The most popular hitsare to be broadcast in a hit parade. The polled population is divided into fourcategories according to sex and age (teenager <= 20, adult> 20). Each person isasked to list five hils, identified by their number between 1 and, say, 50. The resultof the poll is presented as a file; each record lists a respondent's name, first name,sex, age, and his choices ordered according to priority. A program is to compute thefollowing dala:
1. A list of hits ordered according to popularity. Each entry consists of the hitnumber and the number of votes il received. Hits not mentioned are omitted.
2. Four separate Iisls with names and first names of all respondents who hadmentioned in first place one of the three hils most popular in lheir calegory.
PROGRAM hitparade(poll ,oulput);CONST n = 50; (.. number of hils .. )TYPE sex = (male, female);
hilno = 1 ., n;query =RECORD
name, firstname: alia;s: sex;age: 0 .. 99;choice: PACKED ARRAY [1..5] OF hilno
END;VAR i,k,max: inleger;
b: boolean;tolal: ARRAY [hilno] OF integer;coun!: ARRAY [sex,boolean,hitno] OF integer;poll: FILE OF query;
PROCEDURE findnames(x: sex; y: boolean);VAR i,i,k: integer;
selection: SET OF hilno;BEGIN selection := [J; reset(poll);
wrileln(" ---------- -------- ------- - -------- ---- ");( .. find 3 hits most frequenlly listed in this group .. )FOR i := 1 TO 3 DOBEGIN max := 0;
FOR j ;= 1 TO n DOIF max < count[x,y,j] THEN
BEGIN max := count[x,y,j]; k := jEND'
selection' := selection + [k]; count[x,y,k] := 0END;(" lisl persons with one of these hils as firsl choice .. )WHILE NOT eof(poll) DOBEGIN
WITH pollt DOIF s = x THENIF (age >= 20) = y THENIF choice[ 1] If\! selection THEN
writeln(" ",name," ",firslname);get(poll)
ENDEND ("findnames") ;
BEGIN resel(poll);FOR i := 1 TO n DO
BEGIN total[i] := 0;
Text processing
counl[male,true,i] := 0; count[fernnle,true,i] := 0;counl[male,false,i] := 0; counl[female,false,i] := 0
END;(0 collect counts .)WHILE NOT eof(poll} DOBEGIN
WITH pollt DOFOR i := , TO 5 DO
BEGIN b := age >= 20; k := choice[i];counl[s,b,I<] := counl[s,b,k] + 1
END;gel(pol/)
END'(. co;"pule lolals .)FOR i := 1 TO n DO
tolal[i] := counl[rnale,lrue,i] + counl[fernale,true,i]+ count[male,false,i] + counl[female.false,i];
page(output);writeln(" reporl on hit popularity poll");writeln(" lisl of hits ordered after popularilY");writeln(" hit frequency");REPEAT max := 0; k := 0;
FOR i := 1 TO n DOIF max < lolal[i] THEN
BEGIN max := lolal[i]; k := iEND;
IF max > 0 THENBEGIN tolal[k] := 0; writeln(k, max)END;
UNTIL max = 0;writeln(" nmnelists separate by sex and age");writeln(" men "); findnames(male,lrue);writeln(" women"); lindnames(female,true);writeln(" boys "); findnames(nlClle,false);writeln(" girls"); findnames(female,false);wrileln(" end of report"}
END.
19
20
6. Recursion
Pascal programs
1. Compute all nl permutations of the integers 1 ... n.
PROGRAM permute(oulpul);CONST n == 4;VAR i: inleger;
a: ARRAY [i ..n] OF integer;
PROCEDURE print;VAR i: integer;
BEGIN FOR i :== 1 TO n DO wrile(a[i]:3);write1n
END (·prinl·) i
PROCEDURE perm(k: inleger);VAR i,x: integer;
BEGINIF k = 1 THEN print ELSEBEGIN perm(k-i);
FOR i := 1 TO k-1 DOBEGIN x := a[i]; a[i] := ark]; ark] := x;
perm(k-1 );x := a[i]; a[i] := ark]; ark] := Xi
ENDEND
END (·perm·) ;
BEGINFOR i := 1 TO n DO a[i] := i;perm(n)
END.
Recursion 21
2. Convert expressions from infix to postfix form. Each expression Is written on aseparate line. Expressions have the following syntax:
expression =term =factor =
term {("+"'''='') term}.factor {"'" factor}.leiter I "(" expression ")".
PROGRAM postfix(input,output);VAR ch: char;PROCEDURE expression;
VAR op: char;PROCEDURE factor;BEGIN IF ch = "(" THEN
BEGIN read(ch); expression; read(ch) (.) .)END ELSEBEGIN wrile(ch); read(ch)END
END (* factor .. ) ;PROCEDURE term;BEGIN factor;
WHILE eh = .... " DOBEGIN read(ch); factor; write( .. •.. )END
END (. term .) ;BEGIN term;
WHILE (ch="+") OR (ch="-") DOBEGIN op ;= cll; read(ch); term; write(op)END
END (. expression .) ;BEGIN
WHILE NOT eof(input) DOBEGIN wrile(" "); read(ch); expression; writeln; readlnEND
END.
22 Pascal· programs
3. Plot Hilbert curves of orders 1 '" n. Plot procedure produces oulpul for theTeklronix 4010 lerminal. Dala are represented as 12-bil bytes: a call of procedurep 12 appends a byle to lhe oulput file.
PROGRAM hilbert(pf,oulpul);CONST n = 4; hO =512;VAR i,h,x,y,xO,YO: inleger;
ce t we, buf: integer;pf: FILE OF integer; (Oplot file O)
PROCEDURE p12(u: inleger);BEGIN buf := buf ° 4096 + u; ce:= cc + 1;
IF cc = 5 THENBEGIN pft := buf; put(pf);
we := we+1; buf := 0; cc := 0;IF we = 31 THEN
BEGIN ph := 0; put(pf); we := 0END
ENDEND ('p12+) ;
PROCEDURE plol;VfiR u,v: integer;
BEGIN u := x DIV 32; v := y DIV 32;p12(40b+v); p12(140b+y-32+v);p 12(40b+u); p12( 100b+x-32+u);
END ("plot") ;
PROCEDURE selplol;BEGIN p12(35b); pialEND;
PROCEDURE slartplo1;BEGIN ec := 0; wc := 0; buf := 0; rewrile(pf)END;
PROCEDURE endplol;BEGIN x := 0; y := 767; selplol; p12(37b);
REPEAT p12(0) UNTIL cc =0END;
PROCEDURE b(i: integer); FORWARD;PROCEDURE c(i: inleger); FORWARD;PROCEDURE d(i: integer); FORWARD;
PROCEDURE a(i: inleger);BEGIN IF i >0 THEN
BEGIN d(i-1); x := x-hi plot;a(i-1); y := y-h; pial;a(i-1); x := x+h; plot;b(i-1)
ENDEND;
PROCEDURE b;BEGIN IF ; >0 THEN
BEGIN c(i-1); Y := y+h; plol;b(i-1); x := x+h; plot;
Recursion
b(i-1); y := y-h; plot;a(i-1 )
ENDEND;
PROCEDURE c;BEGIN IF i ) 0 THEN
BEGIN b(i-1); X := x+h; plot;c(i-1); y := y+h; plot;c(i-1); X := x-h; plot;d(i-1 )
ENDEND;
PROCEDURE d;BEGIN IF i ) 0 THEN
BEGIN a(i-1); y := y-h; plot;d(i-1); X := x-h; plot;d(i-1); y := y+h; plot;c(i-1 )
ENDEND;
BEGIN slarlplot;i := 0; h := hO; xO := h DIV 2 + h; yO := h DIV 2;REPEAT (·plot hilbert curve OF order j*)
i := i+ 1; h := h DIV 2;xO := xO + (h DIV 2); yO := yO ... (h DIV 2);x := xO; y := yO; setplot;a(i)
UNTIL i = n;endplot
END.
23
24 Pascal programs
4. Plot Sierpinski space-filling curves using their recursive pallern. Plot routine isidentical to lhe one used in the preceding program.
PROGRAM sierpinski(pf,oulpul);CONST n = 4; hO = 512;YAR i,h,x,Y,xO,YO: integer;
CC, wc, bur: integer;pr: FILE OF integer; eplot file·)
PROCEDURE p12(u: integer);BEGIN buf := buf • 4096 + u; ec:= cc + 1;
IF cc = 5 THENBEGIN ph := bur; put(pf);
we := wc+1; buf := 0; cc := 0;IF we = 31 THEN
BEGIN pf t := 0; put(pf); we := 0END
ENDEND (*p12·) ;
PROCEDURE plot;YAR u,v: integer;
BEGIN u := x DIV 32; v := y DIY 32;p12(40b+v); p12(140b+y-32·v);p12(40b+u); p12(100b+x-32·u);
END (·pIOI·) ;
PROCEDURE setplot;BEGIN p12(35b); plotEND;
PROCEDURE startplot;BEGIN ee := 0; wc := 0; bur := 0; rewrite(pf)END;
PROCEDURE endpial;BEGIN x := 0; y := 767; selplol; p12(37b);
REPEAT p12(0) UNTIL cc =0END;
PROCEDURE b(i: integer); FORWARD;PROCEDURE c(i: integer); FORWARD;PROCEDURE d(i: integer); FORWARD;
PROCEDURE a(i: integer);BEGIN IF i ) 0 THEN
BEGIN a(i-1); x := x+h; y := y-h; plot;b(i-1); x := x + 2·h; pial;d(i-1); x := x+h; y := y+h; plot;a(i-1)
ENDEND;
PROCEDURE b;BEGIN IF i ) 0 THEN
BEGIN b(i--1); x := x-h; y := y-h; plot;c(i-1); y := Y - 2·11; plot;
Recursion
a(i-1); X := x+h; y ::: y-h; plot;b(i-1)
ENDEND;
PROCEDURE c;BEGIN IF i > 0 THEN
BEGIN c(i-1); X := x-h; y ::: y+h; plol;d(i-1); X := X - 2 +11; plot;b(i-1); X := x-h; y := y-h; plol;c(i-1 )
ENDEND;
PROCEDURE d;BEGIN IF j > 0 THEN
BEGIN d(i-1); X := x+h; y := y+hi plot;a(i-1); y := y + 2+11; plot;c(i-1); X := x-h; y := y+h; plOI;d(i-1 )
ENDEND;
BEGIN slarlplol;i := 0; h := hO DIV 4; xO ::: 2+h; yO := 3+h;REPEAT i := i+ 1; xO := xO-h;
h := 11 DIV 2; yO ::: yO+h;x := xO; y := yO; selplol;a(i); x := x+h; y := y-h; plot;b(i); X ::: x-h; y := y-h; plot;c(i); x := x-h; y := y+h; plot;d(i); X := x+h; y := y+h; plot;
UNTIL i = n;endpJot
END.
25
26
7. Sorting arrays
Pascal programs
Reference: N.Wirlh, Algorithms + Data Structures =Programs, Prentice-Hall, Inc., 1975
PROGRAM sort(output);CONST n =256; nn = 512;TYPE index = O..nn;
item =RECORD key: integer;(tother fields defined here·)
END;
VAR i : index; r: integer;a: ARRAY [-15..nn] OF item;z: ARRAY [1 ..n] OF integer;
PROCEDURE teste t: alia; PROCEDURE sort);VAR i,z: integer;
BEGIN write(" ", 0;FOR i := 1 TO n DO a[i].key := i;z := clock; sort; write( clock-z);FOR i := 1 TO n DO a[i].key := z[i];z := clock; sort; write(clock-z);FOR i := 1 TO n DO a[i].key := n-i;z := clock; sort; writeln(clock-z);
END (·testO) ;
PROCEDURE slraightinsertion;VAR i,j: index; x: item;
BEGINFOR i := 2 TO n DOBEGIN x := a[i]; e[O] := x; j := i-1;
WHILE x.l~ey <a[j].l<ey DOBEGIN a[j+1) := a[j]; j := j-1;END'
a[j+1] :~ xEND
END;
PROCEDURE binaryinserlion;VAR i,j,I,r,m: index; x: item;
BEGINFOR i := 2 TO n DOBEGIN x := a[i); I := 1; r := i-1;
WHILE I <= r DOBEGIN m := (I+r) DIV 2;
IF x.key < a[m]'key THEN r := m-1 ELSE I := m+ 1;END;FOR j := j-1 DOWNTO I DO a[j+1] := a[j];a[l] := x;
ENDEND;
Sorting arrays
PROCEDURE shellsort;CONST t = 4;VAR i,j,k,s: index; x: item; m: 1..t;
h: ARRAY [1..1] OF integer;BEGIN h[ 1] := 9; h[2] := 5; h[3] := 3; h[4] := 1;
FOR rn := 1 TO t DOBEGIN k := hem]; s := -k; (·sentinel position·)
FOR i := k+ 1 TO n DOBEGIN x := a[i]; j := i-k;
IF s=O THEN s := -k; s := s+l; a[8] := x;WHILE x.key < a[j].key DOBEGIN a[j+k] := a[j]; j:= j-kEND;a[j+k] := x
ENDEND;
END ;
PROCEDURE slraightselection;VAR i,j,k: index; x: item;
BEGIN FOR i := 1 TO n-1 DOBEGIN k := i; x:= a[i];
FOR j := i+1 TO n DOIF a[j].key < x.key THENBEGIN k := j; x := a[j]END;
ark] := a[i]; a[i] := x;END
END;
PROCEDURE heapsort;VAR I,r: index; x: item;
PROCEDURE sift;LABEl 13;VAR i,j: index;
BEGIN i := I; i :0= 2'"i; x := a[i];WHILE j <= r DOBEGIN IF j < r THEN
IF a[j].key < a[i+1 ].key THEN j := i+ 1;IF x.key >= a[j].key THEN GOTO 13;a[i] := a[iJ; i := j; j := 2 t j
END;13: a[i] := xEND;
BEGIN I := (n DIV 2) + 1; r := n;WHILE I > 1 DO
BEGIN I := 1-1; sir!END;
WHILE r ) 1 DOBEGIN x := a[I]; a[l] := a[rJ; a[r] := x;
r := r-1; sirtEND
END ('"heapsorl'") ;
27
28 Pascal . programs
PROCEDURE bubblesorl;VAR i,j: index; x: item;
BEGIN FOR i := 2 TO n DOBEGIN FOR j := n DOWNTO i 00
IF a[j-1 ].key >a[j].key THENBEGIN x ::::: a[j-1]; a[j-1] ::::: a[j); a[j] := Xj
END;END
END (+bubblesorl+) ;
PROCEDURE bubblex;VAR j,k,l: index; x: item;
BEGIN I := 2;REPEAT k := n;
FOR j := n DOWNTO , DOIF 0[j-1 ].key > a[n.key THEN
BEGIN x := a[j-1]; a[j-1] := a[j}; a[j] := x;k:= j
END;I := k+1
UNTIL' >nEND (+bubblex+) ;
PROCEDURE shakersorl;VAR j,k,l,r: index; x:ilem;
BEGIN I := 2; r := n; k := n;REPEAT
FOR j := r DOWNTO , DOIF a[j-1 ].key > a[j].key THENBEGIN x := a[j-1]; a[j-1] := a[j); aD] := x;
k := jEND;
, := k+1;FOR j := I TO r DO
IF a[j-1 ].key > a[j].key THENBEGIN x := a[j-1]; a[j-1] := a[j); aD] := x;
k := jEND;
r := k-1;UNTIL I > r
END (+shakersorl*) ;
Sorting arrays
PROCEDURE quicksort; (·recursive·)
PROCEDURE sort(I,r: index);VAR i.j: index; x,w: item;
BEGIN i := I; j := r;x := a[(I+r) DIV 2];REPEAT
WHILE a[i].key < x.key DO i := i+1;WHILE x.key <a[j].key DO j := j- I ;IF i <= j THENBEGIN w := a[i]; a[i] := a[j]; a[j] := w;
i := i+1; j := j-IEND
UNTIL i > j;IF I < j THEN sorl(I,j);IF i < r THEN sorl(i,r)
END;
BEGIN sort(1,n)END (·quicksort·) ;
PROCEDURE quicksort1; (.non-recursive·)CONST m = 12;VAR i,j,I,r: index;
x,w: item;s: 0 .. m;stack: ARRAY [1 ..m] OF
RECORD l,r: index END;BEGIN s := "; stack[ 1].1 := 1; stack[ 1].r := n;
REPEAT (·tal~e lop request from stack·)I := stack[s].l; r := stack[s].r; s := s-I;REPEAT (·split a[l] ." a[r]·)
i := I; j := r; x := a[(IH) DIV 2];REPEAT
WHILE a[i].key < x.key DO i := i+";WHILE x.key < a[j].key DO j := j-1;IF i <= j THENBEGIN w := a[i]; a[i] := a[j]; an] := w;
i := 1+1; j := j-1END
UNTIL i > j;IF i < r THENBEGIN (·slack request 10 sort right partition·)
s := s+1; stack[s].1 := i; stacl~[s].r := rEND;r := j
UNTIL I )= rUNTIL s = 0
END (·quicksort"'·) ;
29
30 Pascal programs
PROCEDURE mergesorl;VAR i,j,k,l,t: index;
h,m,p,q,r: integer; up: boolean;('nole that a has indices 1..2·n·)
BEGIN up := true; p := 1;REPEAT h := 1; m := n;
IF up THENBEGIN i := 1; j := n; k := n+1; 1:= 2"nEND ELSEBEGIN k := 1; I := II; i := n+1; j := 2'nEND·REPEAT ("merge a run from i ~md j to k")
('q = length of i-run, r = length of j-run")IF m )= p THEN q := p ELSE q := m; m:= m-q;IF m ): p THEN r := p ELSE r := m; m:: m-r';WHILE (q#O) AND (r#O) DOBEGIN ("merge")
IF a[i].key <a[j].key THENBEGIN a[k] := a[i]; k := k+h; i := i+1;'q :: q-1;END ELSEBEGIN a[k] ::: a[j]; k := k+h; j := j-1; r := r-1;END
END;IF q = 0 THEN
BEGIN ('copy tail of j-run")WHILE r # 0 DOBEGIN a[k] := a[j]; k := k+h; j := j-1; r := r-1;END
END ELSEBEGIN (+r = 0, copy tail of i-run")
WHILE q # 0 DOBEGIN ark] := a[i]; k := k+h; i := i+1; q := q-1;END
END;h := -hi t := k; k := I; I := t
UNTIL m = 0;up := NOT up; p := 2"p
UNTIL p )= n;IF NOT up THENFOR i := 1 TO n DO a[i] := a[i+n]
END (+mergesort") ;
BEGIN i := 0; r := 54321;REPEAT i := i+1;
r := (131071+r) MOD 2147483647; z[i] := rUNTIL i :: ,n;test("slr insert", straightinsertion);test("bin insert", binaryinsertion);lesl("shell sort", shellsort);lesl("slr selecl", straightselection);lest("heapsort ", heapsart);lesl("bubblesort", bubblesort);test("bubblesort", bubblex);test("shal(ersorl", shakersorl);test("Quicksorl ", qUicksort);test("quicksorU", quicksor(1);lest("mergesort ", mergesort);
END.
Sequentlal sorting
8. Sequential sorting
1. Natural merge sort wilh 3 files (lapes) and 2 phases.
PROGRAM mergesort(input,output);TYPE item = RECORD key: inleger
(tother fields defined here t )END;
lape = FILE OF item;VAR C: lape; n: bur: item;
PROCEDURE Iist(VAR f: lape);VAR x: ilem;
BEGIN reset(f);WHILE NOT earn) DO
BEGIN read(r,x); wrile(output, x.key: 4)END; .
wrilelnEND ('Iist t ) ;
PROCEDURE n£lturalmerge;VAR I: integer; (tno. at runs mergedO)
ear: boolean; (Oend-ot-run indicalorO)a,b: lape;
PROCEDURE copy(VAR x,y: tape);VAR bur: item;
BEGIN reacl(x, but); wrile(y,but);IF eor(x) THEN ear := true ELSE ear := buf.key > xt .key
END;
PROCEDURE copyrun(VAR x,y: tape);BEGIN (Ocopy one rllll from x 10 yO)
REPEAT copy(x,y) UNTIL earEND;
PROCEDURE dislribute;BEGIN ('rrom c 10 a and b')
REPEAT copyrun(c,a);IF NOT eot(c) THEN copyrull(c,b)
UNTIL eor(c)END;
PROCEDURE mergerun;BEGIN ('from a and b to ct)
REPEATIF at .key ( bt .key. THENBEGIN copy(a,c);
IF ear THEN copyrun(b,c)END ELSEBEGIN copy(b,c);
IF ear THEN copyrun(a,c)END
UNTIL earEND;
PROCEDURE merge;BEGIN ('from a and b to c')
31
32 Pascal programs
REPEATmergerun; I := 1+1UNTIL eof(a) OR eof(b);WHILE NOT eof(a) DOBEGIN copyrun(a,c); I := 1+1END;WHILE NOT eof(b) DOBEGIN copyrun(b,c); I := 1+1END;list(c)
END;
BEGINREPEAT rewrite(a); rewrile(b); reset(c);
dislribute;resel(a); reset(b); rewrite(c);I ;= 0; merge;
UNTIL I = 1END;
BEGIN ('main program. read input sequence ending with 0')rewrite(c); read(buf.key);REPEAT wrile(c, buf); read(buLkey)UNTIL buLkey =0;Iist(c);naturalmerge;Iist(c)
END.
Sequential sorting 33
2. Sequential sorting by n-way mergesorl. In each phase, data are merged from n/2files and distributed onto the other n/2 files. The program starls with the generationof a single file with random numbers.
PROGRAM balancedrnerge(output);CONST n = 6; nh = 3; (·no. of tapes')TYPE item = RECORD
key: integerEND'
tape = I=ILE OF item;tapeno = 1..n;
VAR leng, rand: integer; ('used to generate file')eat: boolean;but: item;fO: tape; ('fO is the input tape with random numbers')f: ARRAY [1..n] OF tape;
PROCEDURE list(VAR t: tape; n: tapeno);VAR z: inleger;
BEGIN wrileln(" tape", n:2); z := 0;WHILE NOT eof(f) DOBEGIN read(f, buf); wrile(outpul, buf.key: 5); z := z+1;
IF z = 25 THENBEGIN writeln(outpul); Z := 0;END;
END;IF z # 0 THEN writeln(output); reset(f)
END (·Iist·) ;
PROCEDURE tapemergesorl;VAR i,i,mx,tx: tapeno;
k 1,k2,1: integer;x, min: integer;I, ta: ARRAY [tapeno] OF tapeno;
BEGIN ('distribute initial runs to t[1] ... t[nh]*)FOR i := 1 TO nil DO rewrile{f[i]);j := nh; I := 0;REPEAT IF j < nh THEN j := j+ 1 ELSE j := 1;
('copy one run from fa to lape j'), := 1+1;REPEAT read(fO, buf); wrile(f[J], buf)UNTIL (buf.key > fat .key) OR eof(fO)
UNTIL eot(fO);FOR i := 1 TO n DO lei] := i;REPEAT ('merge from 1[1] ... t[nll] to t[nh+1] ... l[n]*)
IF I < nh THEN k1 := I ELSE k1 := nh;('k1 = no. of input tapes in this phase')FOR i := 1 TO k1 DO
BEGIN resel(f[t[i]]); list(f[t[i]], t[i]); ta[i] := t[i]END;
f := 0; (" = number of runs merged")j := nll+ 1; ("j == index of output tape')REPEAT ('merge a run from 1[1] '0' l[k1] TO L[j]')
k2 ;= k1; 1:== 1+1; ('k2 = no. of active input tapes')REPEAT ('select minimal element')
i :== 1; mx := 1; min := f[La[1]]t.key;WHILE: i <k2 DOBEGIN i := i+1; x := f[ta[i]]t.key;
34 Pascal· programs
IF X <min THENBEGIN min := x; mx:= iEND
END;('ta[mx] has minimal element, move it to t[j]')read(f[ta[mx]J, bur); eat := eof(f[ta[mxJ]);wrile(f[l[j]], but);IF eot THENBEGIN rewrile(f[ta[mx]]); ('eliminate tape·)
ta[mx] := ta[k2]; ta[k2] := ta[kl];kl := kl-l; k2 := k2-1
END ELSEIF buf.key > f[ta[mx]]t .key THENBEGIN tx := ta[mx]; ta[mx] := ta[k2J; ta[k2J := Ix;
k2 := k2-1END
UNTIL k2 = 0;IF j < n THEN j := j+l ELSE j:= nh+1
UNTIL k1 = 0;FOR i := 1 TO nh DO
BEGIN tx := lei]; t[i] := t[i+nh]; l[i+nh] := txEND
UNTIL I = 1;reset(f[1[1]]); list(f[I[lJ], l[lJ); ('sorled oulput is on t[l]·)
END ('lapemergesort') ;
BEGIN ('generate random fjle fO')leng := 200; rand := 7789; rewrile(fO);REPEAT rand := (131071'rand) MOD 2147483647;
buLkey := rand DIV 2147484; write(fO, buf); leng := leng - 1UNTIL leng =0;reset(fO); Iisl(fO,l);tapemergesort;
END.
Sequential sorling 35
3. Polyphase sort program. There are n-1 source files for merging and a singleoutput file. The destination of the merged data changes, when a certain number ofruns has been distributed. This number is computed according to a Fibonaccidistribution.
PROGRAM polysorl(autput);CONST n = 6; ('no. of tapes')TYPE item = RECORD
key: integerEND;
tape = FILE OF Hem;tapeno = l ..n;
VAR leng, rand: integer; ('used 10 generate file')eat: boolean;buf: item;fO: tape; ('fO is the input tape with random numbers')f: ARRAY [1..n] OF tape;
PROCEDURE Iist(VAR f: tape; n:· tapeno);VAR z: integer;
BEGIN z := 0;writeln(" tape", n:2);WHILE NOT eof( f) DOBEGIN read(f, buf); write(autput, buf.key: 5); z := z+ 1;
IF z = 25 THENBEGIN writeln(autptlt); z := 0END;
END;IF z # 0 THEN writeln(autput}; reset(f)
END ('list') ;
PROCEDURE palyphasesart;VAR i,j,mx,tn: tapeno;
k, level: inleger;a, d: ARRAY [tapeno] OF integer;
('a[i] = ideal number of runs on tape j')('d[n = number of dummy runs on tape j')
dn, x, min, z: integer;last: ARRAY [tapena] OF integer;
('last[iJ = key of tail item on lape it)t,ta: ARRAY [tapena] OF tapena;
('mappings af tape numbers')
PROCEDURE selecttape;VAR i: tapena; z: integer;
BEGINIF d[j] < d[j+ 1] THEN j := j+ 1 ELSEBEGIN IF d[;] = 0 THEN
BEGIN level := level + 1; z := a[l];FOR i := 1 TO n-1 DOBEGIN d[i] := z + a[i+1] - a[i]; a[i] := z + a[i+1]END
END;j := 1
END;d[n := d[j] -1
END;
36 Pascal programs
PROCEDURE copyrun;BEGIN (·copy one run Irom 10 to tape j")
REPEAT read(IO, bur); write(l[iJ, bul);UNTil eol(fO) OR (buLkey ) fOt .key);lasl[j] := buLkey
END;
BEGIN (·distribute initial runs")FOR i := 1 TO n-1 DO
BEGIN a[i] := 1; d[i] := 1;' rewrile( I[i])END;
level := 1; j := 1; a[n] := 0; den] := 0;REPEAT selecllape; copyrunUNTIL eol(lO) OR (j=n-1);REPEAT selecllape;
IF lasl[j] <= rot.key THENBEGIN ("continue old run")
copyrun;IF eol(fO) THEN d[j] := d[j] + 1 ELSE copyrun
ENDELSE copyrun
UNTIL eol(lO);FOR i := 1 TO n-1 DO reset(f[i]);
FOR i := 1 TO n DO t[i] := i;REPEAT ('merge from t[1] ... l[n-1] to t[n]*)
z := a[n-1]; den] := 0; rewrite(l[t[n]]);writeln(" level", level:4," lape", l[n]:4);FOR i := 1 TO n DO wrileln(l[i], a[i], d[i]);REPEAT k := 0; ("merge one run·)
FOR i := 1 TO n-1 DOIF d[i] >0 THEN d[i] := d[i]-1 ELSE
BEGIN k := k+1; taCk] := I[i]END;
IF k = 0 THEN den] := den] + 1 ELSEBEGIN ("merge one real run Irom t[1] ... t[k]")
REPEAT i:= 'I; mx:= 1;min := f[ta[ 1]] t .I<ey;WHILE i < 1< DOBEGIN i := i+1; x := I[ta[i]]t.key;
IF x <min THENBEGIN min := x; mx := iEND
END'("ICi[';'X] contains minimal elemenl, move it 10 I[n] ')read(f[ta[mx]]. 'bul); eot := eol(l[ta[mx]]);wrile(f[t[n]], but);IF (buLkey ) I[ta[mx]]t.key) OR eot THENBEGIN (·drop this tape')
ta[mx] := taCk]; k := k-1END
UNTil k = aEND;z := z-1
UNTil z =0;reset(f[l[n]]); list(f[i[n]], ten]); ("rotate tapes")tn := I[n]; 'dn := den]; z := a[n-1];FOR i := n DOWNTO 2 DO
Sequential sorting
BEGIN I[i] := t[i-1]; d[i] := d[i-1]; a[i] := a[i-1] - zEND;
I[ 1] := tn; d[1] := dn; a[ 1] := z;(. sorted outpul is on I[1] .)level := level - 1
UNTIL level =0;END (·polyphasesorI O
) ;
BEGIN (0 generale random file·)leng := 200; rand := 7789;REPEAT rand:= (131071°rand) MOD 2147483647;
buLkey := rand DIV 2147484; wrile(fO, but); leng := leng UNTIL leng = 0;reset(fO); lisl(fO,1);polyphasesort;
END.
37
38 Pascal programs
9. "Problem solving", backtracking.
1. Find all sellings of 8 queens on an 8x8 chess board such that no queen checksanother queen. [see also, Comm. ACM 14, 4, 22·1-27 (April 74)].
PROGRAM eightqueens(output);VAR i : integer;
a : ARRAY [ 1..8 ] OF boolean;b: ARRAY [2.. 16] OF boolean;c : ARRAY [-7..7 ] OF boolean;x : ARRAY [ 1..8 ] OF integer;safe : boolean;
PROCEDURE print;VAR k: integer;
BEGIN write(" It);FOR k := 1 TO 8 DO write(x[k]:2);wrlteln
END;
PROCEDURE trycol(j : integer);\fAR i : integer;
PROCEDURE setqueen;BEGIN a[i] := false; b[i+j] := false; c[i-j] := falseEND;
PROCEDURE remove queen;BEGIN a[i] := true; b[i+j] := true; c[i-j] := trueEND;
BEGIN i := 0;REPEAT i := i+·l; safe := a[i] AND b[i+iJ AND c[i-j];
IF safe THENBEGIN sctqueen; x[j] := i;
IF j < 8 THEN trycol(j+1) ELSE print;removequeen
ENDUNTIL i = 8
END;
BEGIN FOR i := 1 TO 8 DO a[i] := true;FOR i := 2 TO 16 DO b[i] := true;FOR i := -7 TO 7 DO c[i] := true;lrycol( I);
END.
"Problem solving", backtracking 39
2. Find sequences of digits 0, 1, 2 and of lenglhs 1 '" 90, such lhal lhey con lain nolwo adjacent subsequences lhat are equal.
PROGRAM sequence012(oulput);CONST maxlenglh = 90;VAR n: inleger;
good: boolean;s: ARRAY [1 ..maxlenglh] OF integer;
PROCEDURE prinlsequence;VAR k: inleger;
BEGIN write(" ");FOR k := 1 TO n DO write(s[k]:1);writeln
END (·printsequence');
PROCEDURE changesequence;BEGIN IF sen] = 3 THEN
BEGIN n := n-1; changesequenceEND ELSE sen] := SlIcc(s[n])
END (. changesequence') ;
PROCEDURE lry;VAR i,l,nhalf: inleger;
BEGIN IF n <= 1 THEN good := true ELSEBEGIN I := 0; nhalf := n DIV 2;
REPEAT 1:= 1+1; i:= 0;(. comp8re tails of length I for equality *)REPEAT good := s[n-i] # s[n-H]; i := i+1UNTIL good OR (bl)
UNTIL NOT good OR (I>=nhalf);END
END ('lry') ;
BEGIN n := 0;REPEAT n := n+1; sen] := 1; lry;
WHILE NOT good DOBEGIN chDngesequence; tryEND;prinlsequence
UNTIL n = maxlenglhEND.
40 Pascal programs
3. Find .the sma,llest positive integer that can be represented as the sum of 10 cubes(integers raised to Ihe third power) in Iwo different ways.
PROGRAM sumofcubes(output);VAR i, ill, II, min, at b, k: integer;
j, sum, pwr: ARRAY [1..200] OF inleger;
(. pwr[I~] = power of k, sum[k]:: p[k] + p[j[k]],j[k] = columnindex of lasl considered candidaIe in row k,ih = rowindex of highest considered row,il = rowindex of least still relevant row .)
BEGIN i := 1; if := 1; ih := 2;j[ 1] ::: 1; pwr[ 1] := 1; sum[ 1] := 2;j[2] := 1; pwr[2] := 8; sum[2] := 9;REPEAT
min := sllm[i]; a := i; b := j[i];(* now gel next sum in row i *)IF j[i] :: i THENBEGIN (* there is none left *) iI := il+1;END ELSEBEGIN IF j[i] :: 1 THEN
BEGIN (* the new min wns from the first column, now adda new row before laldng th~ new sum from Ihe old row')ih := ih + 1; pwr[ih] := ill'ih 'ih;j[ih] :=1; sum[ih] := pwr[ih]+1;
END;j[i] := j[i]+1; (. next candidate in row i .)sum[i] := pwr[i] + pwr[j[i]]
END;(* nOli\( find minimal candidate in rows iI .. ih *)i := it; k := i+ 1;WHILE k <= ih DOBEGIN IF sum[k] <sum[i] THEN i := k; k:= k+1END
UNTIL sum[i] :: min;writeln(min,a,b,i,j[i])
END.
"Problem solving". backtracking
4. Find a path of a knight on a chess board which covers all 64 squares.
PROGRAM knightslour(output);CONST n = 5; nsq = 25;TYPE index = 1..n;VAR i,i: index;
q: boolean;s: SET OF index;a,b: ARRAY [1..8] OF integer;h: ARRAY [index, index] OF integer;
PROCEDURE try(i: integer; x,y: index; VAR q: boolean);VAR k,u,v: integer; q1: boolean;
BEGIN k := 0;REPEAT k := k+ 1; q1 := false;
U := x + ark]; v := y + b[k];IF (u IN s) AND (v IN s) THENIF h[u,v] = 0 THENBEGIN h[u,v] := i;
IF i < nsq THENBEGIN try(i+1,u,v,q1);
IF NOT q1 THEN h[u,v] := 0END ELSE q 1 := true
ENDUNTIL q1 OR (1<=8);q := ql
END (>try» ;
BEGIN s := [1,2.3.4,5];8[1]:= 2; b[ 1]:= 1;a[2]:= 1; b[2]:= 2;a[3] := - t; b[3] :=2;a[4] := -2; b[4]:= 1;0[5] := -2; beG] := -1;a[6] := -1; b[6] := -2;a[7]:= 1; b[?] := -2;n[8]:= 2; b[a] := -1;FOR i := 1 TO n DO
FOR j := 1 TO n DO h[i,j] := 0;h[1,1] := 'f; try(2,1,I,q);IF q THEN
FOR i := 1 TO n DOBEGIN FOR j := 1 TO n DO write{h[i,j):5);
write InEND
ELSE writeln(It no solution It)END.
41
42 Pascal programs
5. Find a solution to the stable marriage problem. n men and n women slate theirpreferences of partners. Find n pairs such Ihat no man would prefer 10 be married toanolher woman who would also prefer him 10 her pmtner. A set of pairs is cnlledstable, if no such cases exist [see also Comm. ACM 14, 7, 486-92 (July 71)].
PROGRAM marriage(inpul,output);CONST n =8;TYPE man = 1..n; woman = 1..n; rank = 1..n;VAR m: man; w: woman; r: rank;
wmr: ARRAY [man, ranl~] OF woman;mwr: ARRAY [woman, rank] OF man;rmw: ARRAY [man, woman] OF rank;rwm: ARR/\Y [woman, man] OF rank;x: ARRAY [man] OF woman;y: ARliAY [woman] OF man;single: ARRAY [woman] OF boolean;
PROCEDURE print;VAR m: man; rm, rw: integer;
BEGIN rm := 0; rw := 0;FOR m := 1 TO n DOBEGIN wrile(x[m]:4);
rm := rm + rmw[m,x[m]]; rw := rw + rwm[x[m],m]END;wri Ie InCrm:8,rw:4);
END ('print") ;
PROCEDURE try(m: man);VAR r: rank; w: woman;
FUNCTION stable: boolean;VAR pm: man; pw: woman;
i, lim: rank; s: boolean;BEGIN s := true; i := 1;
WHILE O<r) AND s DOBEGIN pw := wmr[m,i]; i := 1+1;
IF NOT single[pw] THEN s := rwm[pw,m] > rwm[pw,y[pw]]END;i := 1; lim := rwm[w,m];WHILE (i<lim) AND s DOBEGIN pm := rnwr[w,i]; i := 1+'1;
IF pm <rn THEN s := rmw[pm,w] > rmw[pm,x[pm]]END;slable := s
END ("test") ;
BEGIN ("Iry")FOR r := 1 TO n DO8EG1[\1 VI := wmr[m,r];
IF single[w] THENIF stable THENBEGIN x[m] := w; yEw] := m; single[w] := false;
IF m < n THEN try(succ(rn» ELSE print;single[w] := true
ENDEND
END rtry*) ;
"Problem solving". backlrocking
BEGIN writeln("1");FOR m := 1 TO n DO
FOR r := 1 TO n DOBEGIN read(wmr[m,r]); rmw[m,wmr[m,r]] := rEND'
FOR w ;= 1 TO n DOFOR r.:= 1 TO n DOBEGIN read(mwr[w,r]); rwm[w,mwr[w.r]] := rEND·
FOR w ;= 1 TO n DO single[w] := Irue;Iry("I)
END.
5 7 1 2 6 8432 3 7 541 8 68514623 73 2 7 4 1 685725 1 368 41 6 7 5 8 4232 5 763 4813 8 457 2615 3 761 2848 6 357 2 1 41 5 6 2 487 38 7 3 241 5 664738"1252 8 5 4 6 3 7 1752 1 864 3741 5 2 368
43
44 Pascal programs
6. Find an opUmal selection of objects from a given set of n objects under a givenconstrain!. Each object is characterised by two properties v (for value) and w (forweight). The optimal selection is the one with the largest sum of values of itsmembers. The constraint is that the sum of their weights musl not surpass a givenlimit fimv. The algorithm is called branch and bound.
PROGRAM selection(inpul,output);CONST n = 10;TYPE index = 1..n;
object = RECORD v,w: integer END;VAR i: index;
a: ARRAY [index] OF object;Iimw, totv, maxv: integer;wi, w2, w3: integer;s, opts: SET OF index;z: ARRAY (boolean] OF char;
PROCEDURE try(i: index; tW,av: integer);VAR av1: integer;
BEGIN (·try inclusion of object i·)IF tw + n[i].w <= limw THENBEGIN s := S + (i];
IF i < 11 THEN try(i+1, tw+a[i].w, av) ELSEIF av > mtlXV THENBEGIN maxv := BV; opts := sEND;
s := S - [i]END;(·now try without object i') av1 := av - a[i].v;IF av1 >mtlxv THENBEGIN IF i < n THEN try(i+1, tw, avi) ELSE
BEGIN maxv := av1; opts := sEND
ENDEND ('try') ;
BEGIN lotv := 0;FOR i := 1 TO n DO
WITH a[i] DOBEGIN read(w,v); totv := totv + vEND;
read(w1,w2,w3);z[true] ;= "'''; z[false] := II ";
write(" weight ");FOR i := 1 TO n DO wriLe(a[i].w:4);writeln; write(" value "); ,FOR i := 1 TO n DO write(a[i].v:4);writeln;REPEAT limw := w 1; m3XV := 0; s := []; opts := [];
try( 1,O,totv);wrile(limw);FOR i := 1 TO n DO write(" ", z[i IN opls]);wrileln; w1 := wi .1- w2 .
UNTIL W"' > w3END.
Lisl and tree structures. poinlers
10. List and tree structures, pointers.
45
1. A procedure search is to locale records with a given key in an ordered list. If thekey is not present, then a new record is 10 be inserted so that the ordering of keys ismaintained. Use a sentinel at the end of the lis!.
PROGRAM Iist(input,output);TYPE ref = tword;
word = RECORD key: integer;count: integer;next: ref
END;VAR k: integer; root, sentinel: ref;
PROCEDURE search(x: integer; VAR root: ref);VAR w1,w2,w3: ref;
BEGIN w2 := root; w 1 := w2 t .next; sentinel t .I~ey := x;WHILE w1 t .key < x DO
BEGIN w2 := w1; w1 := w2t.nextEND'
IF (w1 t .'key = x) AND (w1 :/I sentinel) THENw 1 t .count := w 1t .count + 1 ELSE
BEGIN new(w3); ('insert w3 between w1 AND w2')WITH w3t DO
BEGIN key := x; count := 1; next := w 1END;
w2t.next := w3END
END ("search+) ;
PROCEDURE printlist(w,z: ref);BEGIN WHILE w :/I z DO
BEGIN wrileln(wt.l~ey. wt.count);w := wt.next
ENDEND (+printlist+) ;
BEGIN new(root); new(senlinel); root t .next := sentinel;read(k);WHILE k:ll 0 DO
BEGIN search(I<, root); read(k)END;
prinllisl(root t .next,senlinel)END.
46 Pascal programs
2. Instead of keeping the list ordered according to keys, reorder it as follows: Aftereach search, the accessed record is moved to the lop of the list. In this case,repeated accesses to the same element will be very fast. Use a sentinel at tile endof the list.
PROGRAM lisl(input,output);TYPE ref = tworu;
word = RECORD key: inleger;count: integer;next: ref
END;VAR k: integer; root, sentinel: ref;
PRO'CEDURE search(x: integer; VAR rool: ref);VAR w1,w2: ref;
BEGIN wi := root; sentinelt.key := x;IF wi = sentinel THENBEGIN (·Iirst element·) new(root);
WITH matt DOBEGIN key := x; count := 1; next := sentinelEND
END ELSEIF wI t .key = x THEN wit .count := wI t .count + 1 ELSEBEGIN (·search·)
REPEAT w2 := wi; wi := w2t.nextUNTIL wIt .key = x;IF wi = sentinel THENBEGIN ('insert·)
w2 := 1'001; new(root);WITH roott DO
BEGIN key := x; count := 1; next := w2END
END ELSEBEGIN (. found, now reorder')
wit.count := wit.counl + 1;w2t.next := wi t.next; wi t.next := root; root := wi
ENDEND
END ('search') ;
PROCEDURE prinllist(w,z: ref);BEGIN WHILE w # z DO
BEGIN writeln(wt .key, wt .count);w := wt.next
ENDEND ('printlist') ;
BEGIN new(sentinel); root := senlinel;read(k);WHILE k # 0 DO
BEGIN search(k, root); read(k)END;
prinllisl(rool,senlinel)END.
List and tree structures, pointers 47
3. Read a sequence of relations defining a directed, finite graph. Then eslablishwhether or nol a partial ordering is defined. If so, print the elements in a sequenceshowing the partial ordering. (Topological sorting).
PROGRAM topsorl(inpul,oulput);TYPE Iref = tleader;
tref = ttrailer;leader = RECORD key: integer;
count: integer;tmil: tref;next: Iref;
END;trailer= RECORD id: Iref;
next: trefEND;
VAR head, tail, p,q: Iref;I: tre!; z: integer;x,y: inleger;
FUNCTION I{w: integer): Iref;('reference to leader with key w')VAR h: Iref;
BEGIN h ;= head; tailt.key := w; (*sentinel·)WHILE h t .key # w DO h:= ht .next;IF h = tail THEN
BEGIN ('no element with key w in the list·)new(tail); z := z+1;ht.counl := 0; ht.trail := NIL; ht.next := lail
END;I := h
END (·1') ;
BEGIN ('initialise list of leaders with a dummy')new(head); tail := head; z := 0;
('input phase') read(x);WHILE x # 0 DOBEGIN read(y); writeln(x,y);
p := I(x); q := I(y);new(l); tt.id ;= q; tt.next := pt.trail;pt.lrail := t; qt.count := qt.count + 1;read(x)
END;
('search for leaders with counl = 0')p := head; head := NIL;WHILE P # lail DOBEGIN q := p; p := pt .next;
IF q t .counl = 0 THENBEGIN qt .nexl := head; head := qEND;
END;
('oulput phase') q:= head;WHILE q /1 NIL DOBEGIN writeln(qt.key); z ;= z-1;
1:= qt.tr<:lil;'q := qt.nexl;WHILE t # NIL DO
48 Pascal -programs
BEGIN p := tt .id; pt .count := pot .count - 1;IF Pt .count = 0 THENBEGIN ('insert pt in q-list+)
p t .next := q; q := pEND;t := tt.next
ENDEND;
IF z # 0 THEN wrileln(" this set is not partially ordered")END.
List and tree structuros. pointers 49
4. Insertion and deletion in a binary tree. Read a sequence of integers. A positiveinteger signifies that it should be inserted in an ordered binary tree as the key of anode. A negative integer signifies that a node with its absolule value as key shouldbe searched and deleted.
PROGRAM tree(input,output);TYPE ref = tword;
word = RECORD key: integer;count: integer;lefl, right: ref;
END;VAR rool: ref; k: integer;
PROCEDURE printlree(w: ref; I: integer);VAF~ i: integer;
BEGIN IF w # NIL THENWITH wt DOBEGIN printlree(left, 1+1);
For-l i := 1 TO I DO write(H ");wrileln(key);printlree(right, 1+1)
ENDEND;
PROCEDURE search(x: integer; VAR p: ref);BEGIN
IF p = NIL THENBEGII'l ("word is not in tree; insert it')
new(p);WITH pt DO
BEGIN key := x; count := 1; left := NIL; right := NILEND
END ELSEIF x < pt.key THEN search(x, pt.lelt) ELSEIF x > pt .key THEN search(x, pt .right) ELSE
pt .count := pt .count + 1END ('search') ;
PROCEDURE delete(x: integer; VAR p: ref);VAR q: ref;
PROCEDURE del(VAR r: ref);BEGIN IF rt .right # NIL THEN del(rt .right) ELSE
BEGIN qt .key := rt .key; qt .count := rt .count;q := r; r := rt .Ielt
ENDEND;
BEGIN ('delete')IF p :: NIL THEN wrileln(" word is not in tree") ELSEIF x < p t.1<ey THEN delete(x, p"t.left) ELSEIF x ) pt.key THEN delete(x, pt.right) ELSEBEGIN ('deleto pt") q:= p;
IF qt .right = NIL THEN P := qt .left ELSEIF qt.left = NIL THEN p:= qt.right ELSE del(qt.loft);('dispose(q)')
ENDEND ('delete') ;
50 Pascal programs
BEGIN rool ;= NIL; read(k);WHILE k # 000BEGIN IF k > 0 THEN
BEGIN writeln(" Insert", k); search(k,root)END ELSEBEGIN wrileln(" delete",-k); delete(-k,root)END;prinllree(rool,O); read(k)
END;END.
List and tree structures. pointers 51
5. Insertion and deletion in a AVL -balanced tree. In the previous program, the binaryIree may grow in all sorls of shapes -- if Ihe inserted keys are ordered upon arrival,the "Iree" even degenerales inlo a linear lis!. In the following program, a balance ismainlained, suell lhat at each node lhe heights of ils Iwo subtrees differ by at most 1.
PROGRAM ballree(inpul,output);TYPE ref = tword;
word = RECORD key: integer;counl: integer;left, right: ref;bal: -1..+1
END;VAR root: ref; h: boolean; k: integer;
PROCEDURE printtree(w: ref; I: integer);VAR i: inleger;
BEGIN IF w # NIL THENWITH wt DOBEGIN prinllree(left, 1+1);
FOR i := 1 TO 1 DO write(" ");wrileln(key:5, bal:3);printtree(right, 1+1)
ENDEND;
PROCEDURE search(x: inleger; VAR p: ref; VAR h: boolean);VAR p" ,p2: ref; ('11 = false')
BEGINIF P = NIL THENBEGIN (·word is not in tree; inserl iI*)
new(p); h := Irue;WITH pt DOBEGIN key := x; count:= 1;
left := NIL; right := NIL; bal := 0END
END ELSEIF x < pt.key THENBEGIN search(x, pt .left, h);
IF h THEN (*Iert branch has grown higher·)CASE p t .bal OF
1: BEGIN pt .bal := 0; h := falseEND;
0: p t .bal := -1;-1: BEGIN (+rebalance+) p1 := pt.left;
IF p1 t .bal = -1 THENBEGIN (*single LL rotalion·)
pt.lefl:= p1t.right; p1t.righl:= p;pt.bal := 0; p := p1
END ELSEBEGIN (+double LR rotation·) p2:= p1 t.right;
p1t.righl:= p2t.lefl; p2t.lell:= p1;p't .left := p2t .right; p2t .rigl1l := p;IF p2t.bal = -1 THEN pt.bal := +1 ELSE pt.bal := 0;IF p2t.bal = +1 THEN p1t.bal:= -I ELSE ph.bal:= 0;p := p2
END;pt .bal := 0; h := false
END
52 Pascal programs
ENDEND ELSEIF X > pt .key THENBEGIN search(x, pt .right, h);
IF h THEN ('right branch has grown higher·)CASE pt .bal OF
-1: BEGIN pt .bal := 0; h := falseEND;
0: pt.bal:= +1;1: BEGIN (·rebalance") pi := pt.right;
IF pH.bal = +1 THENBEGIN ("single RR rotation·)
pt.rigl1t := pH.left; pi t.lefl := p;pt .bal := 0; p := pi
END ELSEBEGIN ("double RL rotation') p2:= pi t .Ieft;
pi t.left := p2t.righl; p2t.right := pi;pt.right:= p2t.left; p2t.left:= p;IF p2t.bal = +1 THEN pt.bal := -1 ELSE pt.bal := 0;IF p2t.bal = -1 THEN pH.bal := +'1 ELSE pH.bal := 0;p := p2
END;pt .bal := 0; h := false
ENDEND
ENDELSEBEGIN pt .count := p t .count ... 1; h := falseEND
END (·semch» ;
PROCEDURE delete(x: integer; VAR p: ref; VAR 11: boolean);VAR q: ref; (+h = false»
PROCEDURE balance1(VAR p: ref; VAR h: boolean);VAR p',p2: ref; b1,b2: -1..+1;
BEGIN (+h = true, left branch has become less high»CASE pt .bal OF
-1: pt .bal := 0;0: BEGIN p t .bal := +1; h := false
END;1: BEGIN (>rebalance') pi := pt.right; b1 := pit.bal;
IF b1 >= 0 THENBEGIN ('single RR rotation·)
pt.right:= pH.left; pit.left:= p;IF b1 = 0 THENBEGIN pt.bal := +1; pi t.bal := -1; h := falseEND ELSEBEGIN P t .hal := 0; pit .bal := 0END;p := pi
END ELSEBEGIN (·double RL rotntion+)
p2 := p'l t.lell; b2 := p2t.bal;pi t.left := p2t.rirJht; p2uight := p'l;pt.right := p2t.lert; p2t.lefl := p;IF b2 = +1 THEN Pt .b81 := -I ELSE Pt .bal := 0;IF b2 = -1 THEN plt.bal := +1 ELSE pit.bal := 0;
list and tree structures, pointers
p := p2; p2t .bal := 0END
ENDEND
END (·balance1·) ;
PROCEDURE balance2(VAR p: ref; VAR h: boolean);VAR p1,p2: ref; b1,b2: -1..+1;
BEGIN (·11 = true, rigl1t branch has become less high·)CASE pt.bal OF
1: pt.bal:= 0;0: BEGIN pt .bal := -1; h := false
END; i
-1: BEGIN (·rebalance·) pi := pt.lelt; b1 := p1t.bal;IF b1 <= 0 THENBEGIN ("single LL rotation·)
p't.left := pi t.right; pi t.rigllt := p;IF b'l = 0 THENBEGIN pt,bal := -1; pi t.bal := +1; h := falseEND ELSEBEGIN pt.bal := 0; p1t.bal := 0END;P := pi
END ELSEBEGIN (·double LR rotation·)
p2 := p 1t .right; b2 := p2t .bal;pi t.rigl1t := p2t,left: p2t.lefl := p1;pt .left := p2t .right; p2t .right := p;IF b2 = -1 THEN pt .bal := +1 ELSE pt .bal := 0;IF b2 = +1 THEN p1 t.bal := -1 ELSE pi t.bal := 0;p := p2; p2t .bal := 0
ENDEND
ENDEND (*balance2") ;
PROCEDURE del(VAR r: ref; VAR 11: boolean);BEGIN (·h = false*)
IF rt .right # NIL THENBEGIN del(rt .right,h); IF h THEN balance2(r,h)END ELSEBEGIN qt .key := rt .key; qt .count := rt .count;
r := rt .Ieft; h := trueEND
END;
BEGIN (·delete·)IF p = NIL THEN
BEGIN writeln(" I<ey is not in tree"); h := falseEND ELSE
IF x < p t .I<ey THENBEGIN delele(x,pt .left,h); IF h THEN balance 1(p,ll)END ELSE
IF x > pt.key THENBEGIN delete{x,pt .right,h); IF 11 THEN balc:tnce2(I1,h)END ELSE
BEGIN (·delete"pt·) q:= p;IF qt .right =NIL THEN
53
54 Pascal . programs
BEGIN P := q t .left; h := trueEND ELSE
IF qt .Iell = NIL THENBEGIN P := qt .right; h := trueEND ELSE
BEGIN del(qt.lefl,h);IF h THEN balance1 (p,h)
END'(tdispose(q)')
ENDEND (tdelete') ;
BEGIN read(k); root := NIL;WHILE k # 0 DOBEGIN IF k >= 0 THEN
BEGIN wrileln(tI inserl", k); search( k,root,h)END ELSEBEGIN writeln(tI deleletl,-k); delete(-k,root,h)END'printt~ee(root.O); read(k)
END;END.
List and tree structures, pointers 55
6. Insert and delete elements in a B-tree of page size 2n. Read a sequence of keys;positive values denote insertion, negative ones deletion. Print the resulting B-treeafter each operation.
PROGRAM Blree(input,output);CONST n = 2; nn = 4; ('page size·)TYPE ref = tpage;
item = RECORD key: integer;p: ref;count: integer;
END;page = RECORD m: O..nn; (·no. of items')
pO: ref;e : ARRAY [1 ..nn] OF item;
END;VAR root, q: ref; x: integer;
h: boolean; u: item;
PROCEDURE prinUree(p: ref; I: integer);VAR i: integer;
BEGIN IF p 1/ NIL THENWITH p"t DO ,BEGIN FOR i := 1 TO I DO write(" ");
FOR i := 'I TO m DO write(e[i].l<ey: 4);wrileln;prinllree(pO,I+1 );FOR i := 1 TO m DO prinUree(e[i].p, 1+1)
ENDEND;
PROCEDURE search(x: integer; a:ref;VAR 11: boolean; VAR v: item);
('search key x on B-tree with root a; if found, increment counter. Otherwiseinsert an item with key x and count 1 in tree. If em item emerges to be passedto a lower level, then assign it to v; h:= "tree a has become higher"·)
VAR k,l,r: integer; q: ref; u: item;
PROCEDURE insert;VAR i: integer; b: ref;
BEGIN ('insert u to the right of at.e[r]·)WITH at DOBEGIN IF m < nn THEN
BEGIN In := m+ 1; h := false;FOR i := m DOWNTO r+2 DO e[i] := e[i-1];o[r+'I] := u
END ELSEBEGIN ('page at is full; split it and assign the emerging
item to v·) new(b);IF r <= n THENBEGIN IF r = n THEN v := u ELSE
BEGIN v := e[n];FOR i := n DOWNTO r+2 DO e[i] := e[i-1];e[r+1J := u
END;FOR i := 1 TO n DO bt.e[i] := at.e[i+n];
END ELSE
56 Pascal programs
BEGIN ('insert u in right page·) r:= r-n; v := e[n+1];FOR i := 1 TO r-1 DO bt.e[i] := at.e[i+n+1];bt .e[r] := u;FOR i := r+1 TO n DO bt.e[i] := at.e[i+n]
END;01 := n; bt.m := n; bt .pO := v.p; v.p := b;
ENDEND ('WITW)
END (·inserP) ;
BEGIN ('search key x on page at; h = false')IF a = NIL THENBEGIN ('item with key x is not in tree') h ::: true;
WITH v DOBEGIN key := x; count := 1; P := NILEND
END ELSEWITH at DOBEGIN I := 1; r := 01; ('binary array search '')
REPEAT k := (I+r) DIV 2;IF x <= e[kj.key THEN r := k-1;IF x >= e[k].key THEN I := k+1;
UNTIL r < I;IF I-r > 1 THENBEGIN ('found·) e[k].count := e[k].count + '1; h := falseEND ELSEBEGIN (' item is not on this page')
IF r = 0 THEN q := pO ELSE q := e[r].p;search(x,q,h,u); IF h THEN insert
ENDEND
END ('search') ;
PROCEDURE delete(x: integer; a: ref; VAR h: boolean);('semet, and delete key x in B-tree a; if a page underflow is necessary,balance with adjacent page if possible, otherwise merge; h:= "page a isundersize"')
VAR i,k,l,r: integer; q: ref;
PROCEDURE underflow(c,a: ref; s: integer; VAR h: boolean);('a:: under!low page, C = ancestor page')VAR b: ref; i,I<,mb,mc: integer;
BEGIN me := ct .01; ('h = true, at.m = n-1")IF s < mc THENBEGIN (*b := page to the right of at) s:= 5+1;
b := ct.e[s].p; mb := bt.m; k := (mb-n+1) DIV 2;(*k = no. of items available on adjacent page b*)at.e[n] := ct.e[s]; at.e[n].p := bt.pO;IF k >0 THENBEGIN ('move k items from b 10 at)
FOR i := 1 TO k-1 DO at .e[i+n] := bt .e[i];c t .e[s] := bt .e[k]; cot .e[s].p := b;bt.pO:= bt.e[k].p; mb:= mb-k;FOR i := 1 TO mb DO bt.e[iJ:= bt.e[i+k];bt.m := mb; at.m := n-I+k; h := false
END ELSEBEGIN ('merge pages a fllld bO)
List and tree structures, pointers
FOR i := 1 TO n DO at.e[i+n] := bt.e[i];FOR i := s TO me-1 DO ct.e[i] := c"t.e[i+1];at.m := nn; ct.rn := mc-1; ('dispose(b)')
ENDEND ELSEBEGIN ('b := page 10 the left of at)
IF s = 1 THEN b := et.pO ELSE b := et,e[s-l].p;mb := bt.m + 1; k := (mb-n) DIV 2;IF k > 0 THENBEGIN ('move k Hems from page b 10 a')
FOR i := n-1 DOWNTO 1 DO at.e[i+k] := at.e[i];at.e[k] := ct.e[s]; at.e[k].p := at.pO; mb := rnb-k;FOR i := k-1 DOWNTO 'I DO at .e[i] := bt .e[i+mb];at.pO := bt.e[mb].p;ct.e[s] := bt.e[mb]; ct.e[s].p := a;bt.rn := mb-1; at.m := n-1+k; h := false
END ELSEBEGIN ('merge pages a and b+)
bt.e[mb] := ct.e[s]; bt.e[mb]'p := ot.pO;FOR i := 1 TO n-1 DO b1' .e[l+rnb] := at .e[i];bt.m := nn; cot.m := mc-1; (+dispose(a)')
ENDEND
END ('underflow') ;
PROCEDURE del(p: ref; VAR h: boolean);VAR q: ref; ('global a,k')
BEGINWITH pt DOBEGIN q := e[m].p;
IF q /t NIL THENBEGIN del(q,h); IF h THEN underflow(p,q,m,h)END ELSEBEGIN P t .e[m].p := at .e[k].p; at .e[k] := pt .e[m];
m := rn-1; h := m<nEND
ENDEND ('del') ;
BEGIN ('delete')IF a = NIL THENBEGIN wrilcln(" key is not in tree"); h := falseEND ELSEWITH at DOBEGIN I := 1; r := m; ('binary array search')
REPEAT I~ := (I+r) DIV 2;IF x <= e[k].l~ey THEN I' := k-1;IF x >= e[k].l~ey THEN' := k+1;
UNTIL I > r;IF 1'=0 THEN q := pO ELSE q := e[r].p;IF 1-1' > I THENBEGIN (' round, now delete e[k)")
IF q = NIL THENBEGIN ('a is a lerminal paget) m:= m-1; h := m<n;
FOR i := k TO m 00 eli] := o[i+ I];END ELSEBEGIN del(q,h); IF h THEN under[\ow(n,Cl,r,h)END
57
58 Pascal programs
END ELSEBEGIN delete(x,q,h);IF h THEN underllow(a,q,r,h)END
ENDEND (·delete·) ;
BEGIN root := NIL; read(x);WHILE x It 0 00BEGIN writeln(" search key", x);
search(x,rool,h,u);IF h THENBEGIN ("inserl new base page") q := root; new(root);
WITH roolt DOBEGIN m := 1; pO := q; e[l] := UEND
END;prinllree(rool,1); read(x)
END;read(x);WHILE x # 000BEGIN wrileln(" delete key", x);
delele(x,rool,h);IF h THENBEGIN (·base page size was reduced·)
IF root t.m = 0 THENBEGIN q := root; root := qt .pO; (·dispose(q)·)END;
END'prillll~ee(root,1); read(x)
ENDEND.
List nnd tree structures, pointers 59
('Iine width of printer')
7. Find lhe optimally structured binary search tree for n keys. Known are the searchfrequencies of the keys, b[i] for key[i], and the frequencies of searches witharguments that are not I~eys (represented in the tree). a[i] is the frequency of anargument lying between key[i-l ] and key[i]. Use Knuth's algorithm, Acta Informatica1, 1, 14-25 (197"\). The following example uses Pascal I<eywords as keys.
PROGRAM optimallree(input,output);CONST n = 31; ("no. of keys")
kin == ,10; ("max keylengtl'")TYPE index = O..n;
alra = PACKED ARRAY [1..kln] OF char;VAR ell: char;
k1, k2: integer;id: aHa; ('identifier or key")buf: ARRAY [1 ..kln] OF char; ("character buffer")key: ARRAY [1 ..n] OF alra;i,j,k: integer;a: ARRAY [1 ..n] OF integer;b: ARRA'( [index] OF integer;p,w: ARRAY [index,index] OF integer;r: ARRAY [index,index] OF index;suma, sumb: integer;
FUNCTION baltree(i,j: index): integer;VAR k: integer;
BEGIN k := (i+j+ 1) DIV 2; r[i,j] := k;IF i >= j THEN baltree := b[k] ELSE
ballree := baltree(i,k-1) + ballree(k,j) + w[i,j]END ("baltree') ;
PROCEDURE opllree;VAR x, min: integer;
i,i.I<,h,m: index;BEGIN ("argument: w, result: p,r')
FORi := 0 TO n DO p[i,i] := w[i,i]; ("width of tree h = 0")FOR i := 0 TO n-1 DO ("width of tree h = 1')BEGIN j:= i+1;
p[i,n := p[i,i] + p[j,j]; r[i,iJ:= jEND;FOR h := 2 TO n DO (. h = width of considered tree .)FOR i := 0 TO n-h DO (. i = lett index of considered tree .)BEGIN j:= i+h; (. j = right index of considered tree ")
III := r[i,j-1]; min:= p[i,rn-1] + p[m,j];FOR k := 111+1 TO r[i-t-1,iJ DOBEGIN x := p[i,k-l] -I- p[I<,j];
IF.x < min THENBEGIN m := k; min:= xEND
END;p[i,j] := min + w[i,j]; r[i,j]:= III
END;END ("opllree') ;
PROCEDURE prinllree;CONST Iw = 120;TYPE ref = tnode;
lineposition "= O..Iw;node = RECORD key: alfa;
60 Pascal programs
pas: lineposition;left, right, link: ref
END'VAR root, curr~nt, next: ref;
q,q1,q2: ref;i: integer;k: integer;u, u 1, u2, u3, u4: lineposition;
FUNCTION tree(i,j: index): ref;VAR p: ref;
BEGIN IF i = j THEN p := NIL ELSEBEGIN new(p);
pt .left := tree(i, r[i,j]-1);pt.pos := trunc((lw-l<ln)*k/(n-1» + (kin DIV 2); k:= k+1;pt .key := key[r[i,j]];pt.right :'" tree(r[i,i], j)
END;tree := p
END;
BEGIN k := 0; root:= tree(O,n);current := root; root t .Iink := NIL;next := NIL;WHILE current # NIL DOBEGIN (*proceed down; first write vertical lines·)
FOR i := 1 TO 3 DOBEGIN u := 0; q := current;
REPEAT u1 := qt.pos;REPEAT write(" "); u := u+ 1UNTIL u = u1;write(":"); u := u+1; q := qt.link
UNTIL q =NIL;writeln
END;(*now print master line; descending from nodes on current listcollect their descendants and form next list')
q := current; u := 0;REPEAT unpack(qt .key, buf, '1);
("center key about pos") i:= kin;WHILE buf[i] = " "DO i:= i-1;u2 := qt.pos - ((i-1) DIV 2); u3 := u2+i;ql := q t .Ieft; q2 := q t .right;IF q1 = NIL THEN u1 := 1I2 ELSE
BEGIN u1 := q1t .pos; ql t .link := next; next := q1END;
IF q2 = NIL THEN u4 := u3 ELSEBEGIN u4 := q2t .pos+ 1; q2t .link := next; next := q2END;
i := 0;WI-IILE u <u1 DO BEGIN writer' "); 1I := u+1 END;WI-IILE u <u2 DO BEGIN wrile("-"); u := u+1 END;WHILE u < u3 00 BEGIN i := i+1; write(blll[i]); II := u+1 END;WHILE u < u4 DO BEGIN write("-"); u := LJ+I END;q := qt .link
UNTIL Cj = NIL;writeln;(*now invert next list AND make it current list')
List and tree slruclures, pointers
current := NIL;WHILE nexl # NIL DO
BEGIN q := next; next := q1' .link;q t .link := current; current := q
ENDEND
END (.printlree') ;
BEGIN (·initialize lable of keys and counlers')key[ 1] := "ARRAY"; key[ 2] := "BEGIN "; key[ 3] := "CASE ";key[ 4] := "CaNST "; key[ 5] := "DIV "; key[ 6] := "DOWNTO ";key[ 7] := "DO "; key[ 8] := "ELSE "; key[ 9] := "END ";key[10]:= "FILE "; key[11.1:= "FOR "; key[12]:= "FUNCTION ";key[13] := "GOTO "; key[14]:= "IF "; key[15] := "IN ',',;.key[16] := "LABEL "; key[17]:= "MOD "; key[18] := "NIL ,key[ 19] := "OF "; key[20]:= "PROCEDURE "; key[21] := "PROGRAM ";l<ey[22] := "RECORD "; key[23] := "REPEAT "; key[24] := "SET ";l<ey[25] := "THEN "; key[26]:= "TO "; key[27] := "TYPE ";l<ey[28] := "UNTIL "; l'\ey[29]:= "VAR "; key[30] := "WHILE ";key[31] := "WITH ";FOR i := 1 TO n DO
BEGIN ali] := 0; b[i]:= aEND'
b[O] := '0; k2 := kIn;("scan inpul text and delermine a and bY)WI-liLE NOT eof(input) DOBEGIN reacl(ell);
IF ell IN ["8" .."Z"] THENBEGIN ('identifier or I<ey') k1 := 0;
REPEAT IF k1 < kin THENBEGIN k1 := k1+1; bul[k1] := ehEND;
read(ell)UNTIL NOT (ell IN ["a" .."z", "0" .. "9"]);IF k1 >= k2 THEN 1<2 := k1 ELSEREPEAT bul[k2] := " "; k2 := k2-1UNTIL k2 = k1;paek(bul,1,id);i := "I; j := n;REPEAT 1< := (i+j) DIV 2;
IF key[k] <= iel THEN i := k+1;IF key[k] )= id THEN j := k-1;
UNTIL i > j;IF key[I<] = id THEN a[I<] := 8[k] + 1 ELSE
BEGIN k := (i+j) DIV 2; b[k] := b[k]+"'END;
END ELSEIF ell = ""'''' THEN
REPEAT read(eh) UNTIL ell = """" ELSEIF eh = "("" THEN
REPEAT read(eh) UNTIL eh = "")"END;writeln(" keys and frequencies of occurrence:");suma := 0; sllmb := b[O];FOR i := 1 TO n DOBEGIN sumo := suma+a[i]; sumb := sumb+b[i];
wrileln(b[i- n, ali], " ", key[i])END;
61
62 Pascal programs
wrileln(b[n]);wrHeln(" . ------");writeln(sumb, suma);
('compute w from a and b t)
FOR i := °TO n DOBEGIN w[i,i] := b[i];
FOR j ::: i+1 TO n DO w[i,j] := w[i,j-1] + a[j] + b[j]END;writeln;write(" average path leng1l1 of balanced tree ::");writeln(baltree(O,n)/w[O,n]:6:3); prinllree;
opllree;write In;write(" average palh length of optimal tree =");writeln(p[O,n]/w[O,n]:6:3); prinllree;
('now consider keys only. selling b :: 0')FOR i := 0 TO n DOBEGIN w[i,i] := 0;
FOR j ::: i+1 TO n DO w[i,j] := w[i,j-1] + a[j]END;opltree;writeln;writeln(" optimal tree considering keys only");printlree
END.
Cross reference generators
11. Cross re"ference generators
63
1. Read a text and generate a cross reference table of nil words. i.e. sequences ofcharacters that b(')~Jin with a leiter and consisl of lellers and digils only. Blanks, endsof lines. and special characlers are considered to be separators. Use a binary tree tostore the words encountered.
PROGRAM crossref(f,oulpul);CONST c1 = 10; (·'ength of words")
c2 = 8; ("numbers per line")c3 = 6; (. digits per number")c4 = 9999; (·max line number")
TYPE alfa = PACKED ARRAY [Lc1] OF char;wordref = tword;
itemref = tHem;word = RECORD key: alfa;
first. last: ilemref;left, right: wordref
END;Hem = PACKED RECORD
Ina: 0 ..9999;next: itemref
END;VAR root: wardref;
k.k1: inleger;n: integer; (·current line number·)id: alta;f: text;a: ARRAY [ , ..c 1] OF char;
PROCEDURE search(VAR w1: wordref);VAf~ w: wardraf; x: itemref;
BEGIN w := w1;IF w = NIL THENBEGIN new(w); new(x);
WITH wtDOBEGIN key := id; left := NIL; right := NIL;
first := x; last := xEND;xt.lno := n; xt.nexl := NIL; w1 := w
END ELSEIF id ( wr.key THEN se<lrch(wt.lefl) ELSEIF id >wr.key THEN search(wt.right) ELSEBEGIN new(x); X1' .Ino := 11; xt .next := NIL;
wt,laslt.next := x; wt.last := xEND
END (·search+) ;
PROCEDURE prinllree(w: wordref);
PROCEDURE printword(w: word);VAn I: inle~Jer; x: iternref;
BEGIN wrile(" ". w.key);x := w.firsl; I := 0;REPEAT IF I = c2 THEN
BEGIN wrileln;I :'= 0; write(" ":c 1+1)
END;
64 Pascnl· programs
1:= 1+1; wrile(xt.lno:c3); X := xt.nextUNTIL X = NIL;writeln
END (·printword·) ;
BEGIN IF w # NIL THENBEGIN printtree(wt .Ieft);
printword(w t); prinltree(w t .right)END
END (Oprinllree·);
BEGIN root := NIL; n := 0; k 1 := c 1;page(output); resel(f);WHILE NOT eof(f) DOBEGIN IF n = c4 THEN n ;= 0;
n := n+1; write(n:c3); (·next line·)write(" tI);WHILE NOT 8010(f) DOBEGIN ('scan non-empty line·)
IF ft IN [tla" .."z"] THENBEGIN k := 0;
REPEAT IF k <c1 THENBEGIN k := k+ 1; a[k]:= ft;END;write(ft); get(f)
UNTIL NOT (ft IN ["a".."z.. ,tlO......g"]);IF k >::: k1 THEN k1 ;= k ELSEREPEAT <.1[k1] := ""; k1 := k1-1UNTIL k1 = k;pack(a,f ,id); search(root)
END ELSEBEGIN ('check for quote or comment·)
IF ft = '''''''' THENREPEAT write(ft); get(f)UNTIL f t = ..... tI. ELSE
IF ft ::: "(" THENREPEAT write(ft); get(f)UNTIL ft = "}tI ;
write(ft); get(f)END
END;writeln; get(f)
END;
page( output); printtree(root);END.
Cross reference generators 65
2. Cross reference generator as above, bul using a hash lable ins lead of a binary tree10 store the words encounlered.
PROGRAM crossref(f,output);LABEL 13;CONST c1 = 10; (Olenglh of words")
c2 = 8; ("numbers per line")c3 = 6; ("digits per number")c4 = 9999; (·max line number")p = 997; ("prime number")free = 0' ";
TYPE index = O..p;Hemref = tilem;word = RECORD key: alfa;
first, last: ilemref;101: index
END;ilem = PACKED RECORD
Ina: 0 ..9999;nexl: Hemref
END;VAR i, top: index;
k,k1: inleger;n: integer; ("current line number")id: alia;f: texI;a: ARRAY [Lc1] OF char;lellers, letdigs: SET OF char;I: ARRAY [O..p] OF word; ("hash tableO)
PROCEDURE search;VAR h,d,i: index;
x: itemref; f: boolean;("global variables: I, iel, lop·)
BEGIN h := orcl(id) DIV 4096 MOD p;('Pascal-60aO defines ord on packed characler array of length 10.
Division is needed because division operules on 48 bits only! ")f := false; d := 1;new(x); xt .1110 := n; x t .next := NIL;REPEAT
IF I[h].key = id THENBEGIN ("found") f := true;
t[h].last t .nexl := x; l[h).last := xEND ELSE -
IF l[h].key :: free THENBEGIf\J ("new enlry") I := true;
WITH I[h] DOBEGIN key := id; firsl := x; last := x; fol := topEND;top := h
END ELSEBEGIN ("collision") h := h+d; d := d+2;
IF h >= P THEN h := h-p;IF d = p THEN
BEGIN wrileln(" table overflow"); GOTO 13END
ENDUNTIL f
66 Poscal programs
END ('search') ;
PROCEDURE printlable;VAR i,i,m: index;
PROCEDURE prinlword(w: word);VAR I: integer; x: i1emref;
BEGIN write(" ", w.key);x := w.lirsl; I := 0;REPEAT IF I = c2 THEN
BEGIN wrileln;I := 0; wrile(" ":c1+1)
END;I := 1+ 1; write(x t .lno:c3); x := x t .next
UNTIL x = NIL;writeln
END ('prinlword') ;
BEGIN i := lop;WHILE i # p DOBEGIN ('scan linked list and find minimal key')
m := i; i := l[i].lol;WHILE j # P DO
BEGIN IF l[j).key < t[mlkey THEN m := j;j := t[j),fol
END;printword(l[m]);IF m # i THEN
BEGIN t[m].key := I[i].key;I[m]'first := t[i].lirst; I[mllasl := t[i].last
END;i := l[i).loJ
ENDEND ('prinllable") ;
BEGIN n := 0; k1 := c1; lop := p; reset(f);FOR i := 0 "TO p DO l[i].l~ey := free;Ie tiers := ["a...."z"]; leldigs := leUers + ["0".."9"];WHILE NOT eof(f) DOBEGIN IF n = c4 THEN n := 0;
n := n+1; write(n:c3); ("next line")wrile(" ");WHILE NOT eoln(f} DOBEGIN ("scan non-emply line')
IF ft IN letters THENBEGIN k := 0;
REPEAT IF k < c1 THENBEGIN k := k+1; a[k]:= ft;END;wrile(ft); gel(f)
UNTIL NOT (It IN leldigs);IF k )= k1 THEN k1 :,,; k ELSEREPEAT a[k1]:=" "; k1:= k1-1UNTIL k1 = k;pack(a,l,id); search;
END ELSEBEGIN ('check lor quole or commenl")
IF rt = .""." THEN
Cross reference generators
REPEAT write(ft); gel(f)UNTIL f t = ...nt" ELSE
IF ft = "(" THENREPEAT wrile(ft); gel(f)UNTIL ft = tt}" ;
write(ft); get(f)END
END;writeln; gel(f)
END;
13: page; prinUableEND.
67
68 Pascal programs
12. Syntax analysis
Skeleton compiler which checks lhe syntax of its input text according to the followinggrammar. Principle is top-down, recursive descent with one symbol lookahead. (seealso N.Wirtll, Algorithms + Data Structures: Programs, Ch. 5, Prentice-Hall, Inc. 1975)
program blocl~ "." .block:: [ "CONST" idenl ":" number ("," ident ":" number) ";"]
[ "VAR" ident ("," ident} ";"]( "PROCEDURE" ident ";" block ";") statement.
slatement : [ ident "::" expression I "CALL" ident I"BEGIN" statement (";" statement} "END" I"IF" condilion "THEN" statement I"WHILE" condition "DO" statement ] ,
condition : "ODD" expression Iexpression (":"1" #"'''<''1'' <:"1" )"'''):'') expression
expression:: ["+"1"-"] term ("+"1"-") term}.term :: factor ("'''1''/'') factor}.faclor:: ident I number I "(" expression ")"
PROGRAM PLO(inpllt,outpul);LABEL 99;
CONST norw = 1 I;
txmax = "' 00;nmax :: 14;81 :: 10;chselsize = 128;
("no. of reserved words")('Iength of identifier table')(·max. no of digits in numbers')
(+Iength of identifiers")("for ASCII character set')
('Iast clleraeter read')("'ast symbol read')
Clast identifier read')("ast number read')
('character cOLlnt')('line length')
TYPE symbol :(nul,idenl,nllmber,plus,minus,times,slash,oddsym,eql,neq,lss,leq,glr,geq,lparen,rparen,comma,semicolon,period,becomes,beginsym,endsym,ifsym,thensym,whilesym,dosym,callsyrn,conslsym,varsym,procsym);
alta = PACKED ARRAY [1..aIJ OF char;object = (conslant,variable,prozedure);
VAR cl1: char;sym: symbol;id: alta;num: inleger;cc: inleger;II: integer;I,k: integer;line: ARRA Y [1 ..81] OF char;a: alta;word: ARRAY [1..norw] OF alra;wsym: ARRAY [1..norw] OF symbol;ssym: AF1RAY [char] OF symbol;Inble: ARRAY [O.. lxmaxJ OF
RECORD name: alra;kind: object
END;
Syntax analysis
PROCEDURE error(n: integer);BEGIN wrileln(" ":cc, "t", n:2); GOTO 99END (*error*) ;
PROCEDURE gelsym;VAR i,Lk: integer;
PROCEDURE getch;BEGIN IF cc = II THEN
BEGIN IF eof(inpllt) THENBEGIN wrile(" program incomplete"); GOTO 99END;
II := 0; cc := 0; wrile(" ");WHILE NOT eoln(input) DO
BEGIN II := 11+1; read(eh); write(eh); line[lI] := cllEND;
writeln; II := 11+1; read(line[II])END;ec := ec+1; eh := line[ee]
END ("gelch") ;
BEGIN ('getsym')WHILE ell = DO geteh;IF ell IN ["a" z"] THENBEGIN ('identifier or reserved word') k:= 0;
REPEAT IF k < al THENBEGIN k := k+1; ark] := ehEND;geteh
UNTIL NOT (ch IN [ ..a......z.. ,..O... .'·g..]);IF k >= kk THEN kl< := k ELSE
. REPEAT a[kk] := .. "; kk := kk-1UNTIL kk = k;
id := a; i := 1; j := norw;REPEAT 1<·:= (i+j) DIV 2;
IF id <= word[k] THEN j := k-1;IF id >= word[k] THEN i := k+ 1
UNTIL i > j;IF i-1 > j THEN sym := wsym[k] ELSE sym := ident
END ELSEIF ell IN ["0".."9"] THENBEGIN ('number') k := 0; num := 0; SYm := number;
r~EPEAT num :=IO'num + (ord(ell)-ord("O"»;1< := k+1; gelcll .
UNTIL NOT (ell IN ["0...."9..]);IF k > nmax THEN error(30)
END ELSEIF eh = ":" THENBEGIN gelch;
IF ell :: "=" THENBEGIN sym := becomes; getchEND ELSE sym := nul;
END ELSEIF eh = "(" THENBEGIN geleh;
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70
IF eh : ":" THENBEGIN sym :: leq; gelehEND ELSE sym := Iss
END ELSEIF eh : ")" THENBEGIN getel1;
IF ell : ":" THENBEGIN sym :: geq; gelchEND ELSE s ym := gtr
END ELSEBEGIN sym := ssym[eh]; gelchEND
END ('getsym") ;
Pascal· programs
PROCEDURE block(lx: inleger);
PROCEDURE enler(k: objecl);BEGIN ('enter ohjecl inlo table')
tx := tx -I- 1;WITH lable[lx] DOBEGIN name :: id; kind :: k;END
END ('enler') ;
FUNCTION position(id: alfa): integer;VAR i: integer;
BEGIN ('find identifier id in lable")lable[O].name := id; i:= tx;WHILE table[i].name It id DO i:= ;-1;position := i
END ('position") ;
PROCEDURE consldeclaration;BEGIN IF sym :: ident THEN
BEGIN getsym;IF sym : eql THENBEGIN getsym;
IF sym = number THENBEGIN enler(eonslant); getsymEND
ELSE error(2)END ELSE error(3)
END ELSE error(4)END ('constdeclaralion") ;
PROCEDUr-lE vardeelaration;BEGIN IF sym = ident THEN
BEGIN enter(vtlriable); getsymEND ELSE errore4)
END ('vardeclmalion") ;
PROCEDURE statement;VAR i: integer;
Syntax analysis
PROCEDURE expression;
PROCEDURE term;
PROCEDURE faclor;VAR i: integer;
BEGINIF sym " idenl THENBEGIN i := position(id);
IF i = 0 THEN error( 11) ELSEIF table[i].kind = prozedure THEN error(21);gelsym
END ELSEIF sym = number THENBEGIN getsymEND ELSEIF sym = Iparen THENBEGIN getsym; expression;
IF sym " rparen THEN getsym ELSE error(22)ENDELSE error(23)
END ('factor") ;
BEGIN ('term') factor;WHILE sym IN [limes,slash] DO
BEGIN gelsym; faclorEND
END ("term") ;
BEGIN ("expression·)IF sym IN [plus,minus] THEN
BEGIN getsym; lermEND ELSE lerm;
WHILE sym IN [plus,minus] DOBEGIN gelsym; termEND
END ("expression·) ;
PROCEDURE condition;BEGIN
IF sym = oddsym THENBEGIN getsym; expressionEND ELSEBEGIN expression;
IF NOT (sym IN [eql,neq,lss,leq,glr,geq]) THENorror(20) ELSE
BEGIN gelsym; expressionEND
ENDEND ('condition') ;
BEGIN (·slatement")IF sym = idenl THENBEGIN i := posilion(id);
IF i = OTJ-IEN error( 11) ELSEIF lable[i].kind II variable THEN error(12);
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72 rascal programs
getsym; IF sym =becomes THEN getsym ELSE errore 13);expression
END ELSEIF sym =callsym THENBEGIN getsym;
IF sym # ident THEN error(14) ELSEBEGIN i := position(id);
IF i = 0 THEN errore 11) ELSEIF table[i].kind It prozedure THEN errore 15);getsyrn
ENDEND ELSEIF sym = ifsym THENBEGIN gelsym; condition;
IF sym = thensym THEN getsyrn ELSE errore 16);statement; .
END ELSEIF sym = beginsym THENBEGIN gelsym; stalemenl;
WHILE sym = semicolon DOBEGIN getsym; slalementEND;
IF sym = endsym THEN gelsym ELSE errore 17)END ELSEIF sym = whilesyrn THENBEGIN gelsym; condition;
IF sym = dosym THEN getsym ELSE errore 18);statement
ENDEND ('statement·) ;
BEGIN ('block')IF sym = conslsym THENBEGIN getsym; constdeclaration;
WHILE sym = comma DOBEGIN gelsyrn; constdeclaralionEND;
IF sym = semicolon THEN getsym ELSE error(5)END;IF sym = varsym THENBEGIN gelsym; vardeclaralion;
WHILE sym = comma DOBEGIN getsym; vardeclaralionEND;
IF sym = semicolon THEN gelsym ELSE error(5)END;WHILE sym = procsym DOBEGIN gelsym;
IF sym = ide"l THENBEGIN enler(prozedure); gelsymEND
ELSE errore 4.);IF sym = semicolon THEN gelsym ELSE error(5);bJocl«tx);
Syntax analysis
IF sym = semicolon THEN getsym ELSE error(5);END;statement
END ("block") ;
BEGIN ("main program")FOR ch := chr(O) TO chr(chsetsize-1) DO ssym[ch] := nul;word[ 1] := "BEGIN "; warde 2] ;= "CALL ";word[ 3] := "CONST "; warde 4] := "DO ";word!: 5] := "END "; warde 6] := "IF ";word[ 7] := "000 "; warde 8] := "PROCEDURE ";warde 9]:= "THEN "; word[10]:= "VAR ";word[ 11] := "WHILE ";wsyrn[ 1] := beginsym; wsym[ 2] := callsym;wsym[ 3] ;= constsym; wsym[ 4] ;= dosym;wsym[ 5] := endsym; wsym[ 6] ~= ifsym;wsym[ 7] := oddsyrn; wsym[ 8] := procsym;wsym[ 9] ;= thensymj wsym[ 10] := varsym;wsym[ 11] := whilesym;ssym["+"] := plus; ssymC"-"]:= minus;ssym[ .. • .. ] := times; ssym["/"]:= slash;ssym["("] ;= Iparen; ssyrn[")"]:= rparen;ssym["="] := eql; ssym[","] := comma;ssym["."] ;= period; ssym["#"]:= neq;ssym["<"] := Iss; ssym[")"] := glr;ssym[";"] := semicolon;paget output);cc := 0; II := 0; eh := II "; kk := al; getsym;block(O);IF sym # period THEN error(9);
99; wrilelnEND.
73
Berichte des Instituts fUr Informatik
* Nr. 1 N. Wirth The Programming Language PASCAL
* Nr. 2 N. Wirth Program development by step-wise refi nement
Nr. 3 P. Uiuchl i Reduktion e1ektrischer Netzwerke und Gauss'sche Elimination
Nr. 4 W. Gander, Numerische Prozeduren IA. Mazzario
* Nr. 5 N. Wirth The Programming Language PASCAL (Revised Report)
* Nr. 6 C.A.R.Hoare, An Axiomatic Definition of the Language PASCALN. Wirth
Nf. 7 W. Gander, Numerische Prozeduren IIA. Mazzario
Nr. 8 E. Engel er,E. Wiedmer, Ein Einb1ick in die Theorie der BerechnungenE. Zachos
* Nr. 9 H.P. Frei Computer Aided Instruction: The Author Language andthe System THALES
* Nr.10 K.V. Nori,U. Ammann,K. Jensen, The Pascal "P' Compiler: Implementation NotesH.H. Nage1i, (Revised Edition)Ch. Jacobi
N. ~Jirth
N. Wirth
Nr.11
* Nr. 12
* Nr. 13Nr.14
* Nr.15
Nr.16
Nr.17
* Nr .18*Nr.19
Nr.20
* Nr. 21
G.I. Ugron, Das Informations-System ELSBETHF.R. LUthi
N. Wirth PASCAL-S: A subset and its Implementation
U. Ammann Code Generati on ina PASCAL Compil er
K.Lieberherr Toward Feasible Solutions of NP-Complete Problems
E. Engeler Structural Relations between Programs and Problems
W. Bucher A contribution to solving large linear problems
N. Wirth Programming languages: what to demand and how toaccess them andProfessor C1everbyte's visit to heaven
MODULA: A 1anguage for modu1 ar multi programmi ng
The use of MODULA andDesign and Implementation of MODULA
E. Wiedmer Exaktes Rechnen mit reel len Zahlen
J.Nievergelt,H.P. Frei, XS-O, a Self-explanatory School Computeret al.
Nr.22
Nr.23
Nr.24
P. Uiuchli
K. Bucher
E. Enge1er
Ein Problem der ganzzah1igen Approximation
Automatisches Zeichnen von Diagrammen
Generalized Galois Theory and its Applicationto Complexity
J.Nievergelt,S· tA '1 T 11' h U fIt t'J Heyde t 1 tes, 'Iodes and Tral s: e 1ng t e ser 0 an n erac lve. r System where he is, what he can do, and how to get to Places
A.C. Shaw On the Specification of Graphic Command Languages andtheir Processors
Nr.25
Nr.26
Nr.27
Nr.28
Nr.29
Nr.30
Nr.31
Nr.32
Nr.33
U. Ammann
E. Zachos
N. Wirth
B.Thurnherr,C.A.Zehnder
A.C. Shaw
E. Engeler
N. Hirth
Error Recovery in Recursive Descent Parsers andRun-time Storage Organization
Kombinatorische Logik und S-Terme
MODULA-2
Global Data Base Aspects, Consequences for the RelationalModel and a Conceptual Scheme Language
Software Specification Languages based on regular Expressions
Algebras and Combinators
A Collection of PASCAL Programs
* out of pri nt
