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Solutions to Survo puzzles

Solutions to Survo puzzles are published on this page.

Solutions (July 6, 2006)

Survo puzzle 1/2006 (15)

Survo puzzle 2/2006 (30)

Survo puzzle 3/2006 (1550)

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Solutions (September 18, 2006)

Survo puzzle 4/2006 (3)

Survo puzzle 5/2006 (55)

Survo puzzle 6/2006 (130)

Survo puzzle 7/2006 (200 after detecting the composer)

Many people fond of classical music know that among composers born in December are Beethoven (1770) and Sibelius (1865). Beethoven (1770) can not be accepted for three reasons:

1. 0 is not permitted,
2. the same number (here 7) can appear only once,
3. the sum of the numbers 1+7+7+0=15 is not equal to 20.

Anybody can make sure that Sibelius (1865) is the right choice as follows:
a Google search birthdays +composers leads immediately to
http://www.classical.net/music/composer/ having a link to List #5: Chronological List Of Composers Sorted by Year of Birth.

Possible birth years are found in Survo, for example, by the command

    COMB P,CUR+1 / P=PARTITIONS,20,4 DISTINCT=1 MAX=9
    Partitions 4 of 20: N[P]=12
    1 2 8 9
    1 3 7 9
    1 4 6 9
    1 4 7 8
    1 5 6 8
    2 3 6 9
    2 3 7 8
    2 4 5 9
    2 4 6 8
    2 5 6 7
    3 4 5 8
    3 4 6 7

and then plausible years (after 1499) are

    1829,1928,
    1739,1793,1937,1973,
    1649,1694,1946,1964,
    1748,1784,1847,1874,
    1568,1586,1658,1685,1856,1865.

By downloading the list mentioned above of 784 more or less famous composers into the edit field of Survo, it is easy to find out that Sibelius is the only alternative.

The hint about December is important since, for example, 1685 is the year for three composers, J.S.Bach, Handel, and Domenico Scarlatti, but none of them is born in December. Without this hint there might be a temptation to try 1685, but in this case there is no solution for this Survo puzzle; and I did not want to cause such a disappointment to anybody.

The solution of 7/2006 is therefore

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Solutions (October 2, 2006)

Survo puzzle 8/2006 (10)

Survo puzzle 9/2006 (100)

Survo puzzle 10/2006 (230)

Survo puzzle 11/2006 (2450)

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Solutions (October 16, 2006)

Survo puzzle 12/2006 (15)

Survo puzzle 13/2006 (80)

Survo puzzle 14/2006 (530)

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Solutions (October 30, 2006)

Survo puzzle 15/2006 (45)

Survo puzzle 16/2006 (300)

Survo puzzle 17/2006 (1200)

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Solutions (November 13, 2006)

Survo puzzle 18/2006 (50)

Survo puzzle 19/2006 (460)

Survo puzzle 20/2006 (1850)

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Solutions (November 27, 2006)

Survo puzzle 21/2006 (30)

Survo puzzle 22/2006 (210)

Survo puzzle 23/2006 (670)

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Solutions (December 11, 2006)

Survo puzzle 24/2006 (0)

Survo puzzle 25/2006 (170)

Survo puzzle 26/2006 (185)

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Solutions (January 2, 2007)

Survo puzzle 27/2006 (13)

Survo puzzle 28/2006 (70)

Survo puzzle 29/2006 (2200)

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Solutions (January 15, 2007)

Survo puzzle 30/2007 (1)

Survo puzzle 31/2007 (40)

Survo puzzle 32/2007 (325)

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Solutions (January 29, 2007)

Survo puzzle 33/2007 (0)

Survo puzzle 34/2007 (100)

Survo puzzle 35/2007 (870)

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Solutions (February 12, 2007)

Survo puzzle 36/2007 (25)

Survo puzzle 37/2007 (110)

Survo puzzle 38/2007 (310)

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Solutions (February 26, 2007)

Survo puzzle 39/2007 (10)

Survo puzzle 40/2007 (55)

Survo puzzle 41/2007 (?)

This is a generalization of the previous problem. It has not a unique solution. List all solutions.

The solution by Survo follows at first the same paths as the procedure by Reijo Sund for finding all essentially different open 3 x 3 Survo puzzles with a unique solution (see Appendix 1 of the expository paper of Survo puzzles).

*All possible 3x3 are listed (N=362880):
*COMB P TO SRIS.TXT / P=PERMUTATIONS,9 RESULTS=1
*
*Copying the text file obtained by COMB to a Survo data file:
*FILE SAVE SRIS.TXT TO NEW SRIS01 / NEWSPACE=80,20
*.......................................................................
*The following general symbols are used in tables:
*
*X1 X2 X3 R1
*X4 X5 X6 R2
*X7 X8 X9 R3
*S1 S2 S3
*
*.......................................................................
*Computing row and columns sums and a combined variable JARJ
*describing the margins uniquely:
*
*VAR R1:1,R2:1,R3:1,S1:1,S2:1,S3:1,JARJ:8 TO SRIS01
*R1=X1+X2+X3
*R2=X4+X5+X6
*R3=X7+X8+X9
*S1=X1+X4+X7
*S2=X2+X5+X8
*S3=X3+X6+X9
*JARJ=S3+100*S2+10000*S1+1000000*R3+100000000*R2+JARJ2
*JARJ2=10000000000*R1+1000000000000
*
*.......................................................................
*
*Computing the JARJ value for the given 3x3 table:
*S1=14 S2=15 S3=16
*R1=12 R2=17 R3=16
*
*JARJ=S3+100*S2+10000*S1+1000000*R3+100000000*R2+JARJ2
*JARJ2=10000000000*R1+1000000000000
*     1000000000000
*JARJ=1121716141516
*
*The "accepted" tables copied to a new Survo data file:
*IND=JARJ,1121716141516
*FILE COPY SRIS01 TO NEW SRISN
*......................................
*Getting the number of oservations:
*STAT SRISN,CUR+1,STATUS
*#var=16 #act=16 #obs=26
*
*Listing all possible 26 solutions (by rows):
*MASK=AAAAAAAAA-------
*FILE LOAD SRISN
*DATA SRISN*,A,B,C
C X1 X2 X3 X4 X5 X6 X7 X8 X9
A  1  2  9  6  8  3  7  5  4
*  1  2  9  8  6  3  5  7  4
*  1  3  8  4  7  6  9  5  2
*  1  4  7  5  9  3  8  2  6
*  1  7  4  8  6  3  5  2  9
*  1  8  3  6  2  9  7  5  4
*  1  8  3  9  2  6  4  5  7
*  2  1  9  5  8  4  7  6  3
*  2  1  9  7  6  4  5  8  3
*  2  4  6  5  3  9  7  8  1
*  2  6  4  7  1  9  5  8  3
*  2  6  4  9  1  7  3  8  5
*  3  2  7  5  4  8  6  9  1
*  4  1  7  2  9  6  8  5  3
*  4  2  6  3  5  9  7  8  1
*  4  2  6  9  5  3  1  8  7
*  4  5  3  2  9  6  8  1  7
*  4  5  3  9  2  6  1  8  7
*  4  6  2  7  1  9  3  8  5    Solution of the problem 40/2007
*  4  7  1  8  3  6  2  5  9
*  5  4  3  1  9  7  8  2  6
*  5  4  3  8  2  7  1  9  6
*  7  1  4  2  6  9  5  8  3
*  7  4  1  5  3  9  2  8  6
*  8  1  3  2  9  6  4  5  7
B  8  3  1  4  7  6  2  5  9
*

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Solutions (March 12, 2007)

Survo puzzle 42/2007 (5)

Survo puzzle 43/2007 (780)

Survo puzzle 44/2007 (?)

Extra requirement: The sum of reciprocal values of numbers in the first row is equal to 1, i.e.
1/A1+1/B1+1/C1+1/D1+1/E1=1
and the numbers in that row are in increasing order.
The row can be revealed in Survo, for example, by a short command of 11 characters.

Thus the number 1 must be presented as a five-term Egyptian fraction and this is solved in Survo simply by the command

EGYPT 1/1,5

EGYPT 1/1,5 / 1/1=1/2+1/4+1/10+1/12+1/15

Another way (used by Risto Nieminen) is to create all permitted partitions of 43 by the COMB command and sums of inverse values as follows:

COMB P TO K.TXT / P=PARTITIONS,43,5 DISTINCT=1 MAX=20
FILE SAVE K.TXT TO NEW K
VAR S:8=1/X1+1/X2+1/X3+1/X4+1/X5 TO K
............................
FILE LOAD -K / IND=S
  2  4 10 12 15       1.000

The solution is

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Solutions (March 26, 2007)

Survo puzzle 45/2007 (5)

Survo puzzle 46/2007 (270)

Survo puzzle 47/2007 (2260)

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Solutions (April 9, 2007)

Survo puzzle 48/2007 (1)

Survo puzzle 49/2007 (220)

Survo puzzle 50/2007 (310)

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Solutions (April 23, 2007)

Survo puzzle 51/2007 (35)

Survo puzzle 52/2007 (530)

Survo puzzle 53/2007 (1100 without the hint)

Hint: Let X be the positive root of the quadratic equation
456*X^2+1416*X-131=0. Then X can be represented as a continued fraction

          1
X = -----------------
              1
    A1 + -----------------
                  1
         B1 + ------------
                      1
              C1 + -------
                   D1 + X

and this gives the the first row of the table.
In Survo these numbers can be evaluated, for example, by the following
recursive formulas of editorial computing:

Y(N,X):=if(N=0)then(X)else(1/(Y(N-1,X)-A(N-1,X)))
A(N,X):=if(N=0)then(int(X))else(int(Y(N,X)))

Solution:
Any real number X being an irrational root of an algebraic equation of second degree
with rational coefficients can be represented as an infinite, periodic
continued fraction. The length of the period may vary very much.

Especially, when 0<X<1 and the length of the period is 4, the representation
has the above form. Since X itself appears as the last item in the last
denominator, the infinite sequence of terms in continued fraction representation
is of the form
A1,B1,C1,D1,A1,B1,C1,D1,A1,B1,...
The standard notation for a continued fraction is
X = [0,A1,B1,C1,D1,A1,B1,C1,D1,...], where 0 refers to the integer part
and a periodic case is typically denoted by
       ___________
X = [0,A1,B1,C1,D1] .

Conversely, it is easy to detect by reducing the above formula of X to
a common fraction that it has the form

    a1 + b1*X
X = -----------,
    a2 + b2*X

indicating that X is a root of an equation of second degree.

...................................

The positive root of the equation given in the hint can be obtained,
for example, by the means of editorial computing in Survo as
X=root_of(456*x^2+1416*x-131,0,1)

and the recursive formulas

Y(N,X):=if(N=0)then(X)else(1/(Y(N-1,X)-A(N-1,X)))
A(N,X):=if(N=0)then(int(X))else(int(Y(N,X)))

give in this case the terms

 A(0,X).=0
 A(1,X).=11
 A(2,X).=8
 A(3,X).=5
 A(4,X).=3
 A(5,X).=11
 A(6,X).=8
 A(7,X).=5
 A(8,X).=3
 A(9,X).=11

The length of the period is obviously 4 and the required numbers are
A1=11, B1=8, C1=5 ja D1=3.

By putting these numbers in the table the final solution of
53/2007 is easily found as

Kimmo Vehkalahti (KV) has developed a somewhat new, rather interesting
method of solving Survo puzzles, based on the binary representation of
the puzzle and operations of Boolean algebra. The matrix computations
of Survo, e.g. Kronecker product, have a central role in the method.

Using this method he has solved Problem 53/2007 without the hint.
The following is a short description of the solution by KV:

.................................................................

I begin by sketching the idea briefly:

First, the possible choices are found for the row 1 and column A,
then similarly for the row 2 and column B. These are combined as
1A and 2B. In the diagrams, the periods symbolize the choices and
the letter 'o' shows their common numbers A1 and B2:

  1A:                 2B:
     A  B  C  D          A  B  C  D
  1  o  .  .  .  27   1  *  .  *  *  27
  2  .  *  *  *  13   2  .  o  .  .  13
  3  .  *  *  *  53   3  *  .  *  *  53
  4  .  *  *  *  43   4  *  .  *  *  43
    47 39 29 21         47 39 29 21

Next, 1A and 2B are combined, introducing two new common choices,
A2 and B1. These are denoted by '+' in the diagram 1A2B:

  1A2B:               3C:                 1A2B3C:
     A  B  C  D          A  B  C  D          A  B  C  D
  1  o  +  .  .  27   1  *  *  .  *  27   1  o  +  +  .  27
  2  +  o  .  .  13   2  *  *  .  *  13   2  +  o  +  .  13
  3  .  .  *  *  53   3  .  .  o  .  53   3  +  +  o  .  53
  4  .  .  *  *  43   4  *  *  .  *  43   4  .  .  .  *  43
    47 39 29 21         47 39 29 21         47 39 29 21

Finally, a similarly formed 3C is combined with the previous ones
leading to the combination 1A2B3C, which includes 6+3=9 common numbers
and covers almost the whole puzzle. The solution follows immediately.

.....................................................................

My implementation in Survo is based on the binary representation
of the puzzle:

COMB P2 TO P53_2.TXT / P2=PARTITIONS,13,4 DISTINCT=1
FILE MAKE X53_2,16,0,X,1
FILE SAVE P53_2.TXT TO X53_2
XALL X53_2,X1,16
TRANSFORM X53_2 BY if(X=MISSING)then(0)else(1)
MAT SAVE DATA X53_2 TO P2
MAT CLABELS NUM(1) TO P2
MAT LOAD P2 11 CUR+2

MATRIX P2
///       1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16
  1       1  1  1  0  0  0  1  0  0  0  0  0  0  0  0  0
  2       1  1  0  1  0  1  0  0  0  0  0  0  0  0  0  0
  3       1  0  1  1  1  0  0  0  0  0  0  0  0  0  0  0

The possible combinations are found by Kronecker product:

MAT DIAG=VEC(IDN(16,16))'
MAT P1A=SUB(KRONECKER(P1,PA),*,DIAG)
MAT DIM P1A /* rowP1A=1647 colP1A=16

Only those ones having exactly one common number are accepted:

FILE SAVE MAT P1A TO P1A / TYPE=1
FILE MASK P1A,CASE,1,-
VARSTAT P1A,SUM:1,SUM
FILE MASK P1A,CASE,1,A
FILE MASK P1A,SUM,1,-
..................
MAT SAVE DATA P1A TO P1A / IND=SUM,1
MAT CLABELS NUM(1) TO P1A
MAT DIM P1A /* rowP1A=809 colP1A=16

The basic choices are combined. Recoding guarantees uniqueness:

MAT PA=2*PA
/PSUM P1A P1 PA

The coding scheme is more detailed to that shown in the beginning:

  1A: (a)             1A: (b)
     A  B  C  D            A    B    C    D
  1  o  .  .  .  27   1  [ 3] [ 1] [ 1] [ 1] 27
  2  .  *  *  *  13   2  [ 2] [ 0] [ 0] [ 0] 13
  3  .  *  *  *  53   3  [ 2] [ 0] [ 0] [ 0] 53
  4  .  *  *  *  43   4  [ 2] [ 0] [ 0] [ 0] 43
    47 39 29 21           47   39   29   21

The combination 2B follows similarly. Both are recoded and then
further combined using Kronecker product again:

MAT TRANSFORM P1A BY 2*X#+1
MAT TRANSFORM P2B BY 2^X#
MAT RLABELS NUM(1) TO P1A
MAT RLABELS NUM(1) TO P2B
MAT P1A2B=SUB(KRONECKER(P1A,P2B),*,DIAG)
MAT DIM P1A2B /* rowP1A2B=99507 colP1A2B=16

There are almost 100 000 combinations, but the conditions will
narrow then down dramatically:

FILE SAVE MAT P1A2B TO P1A2B / TYPE=1
FILE MASK P1A2B,CASE,1,-
VARSTAT P1A2B,z1:1,#VAL,6
VARSTAT P1A2B,z2:1,#VAL,14,56
VARSTAT P1A2B,z3:1,#VAL,7
VARSTAT P1A2B,z4:1,#VAL,8
VARSTAT P1A2B,z5:1,#VAL,10
VARSTAT P1A2B,z6:1,#VAL,12
..........................
VAR OK=1 TO P1A2B / SELECT=a*b*c*d*e*f
a=z1,0 b=z2,0 c=z3,1 d=z4,1 e=z5,1 f=z6,1
.................
FILE MASK P1A2B,CASE,1,A
FILE MASK P1A2B,z1,1,-...
.............................
MAT SAVE DATA P1A2B TO P1A2B / IND=OK
MAT CLABELS NUM(1) TO P1A2B
MAT DIM P1A2B /* rowP1A2B=7 colP1A2B=16

Only seven choices are left. Now, the extended coding scheme
covers almost the whole puzzle:

  1A2B: (a)           1A2B: (b)
     A  B  C  D            A    B    C    D
  1  o  +  .  .  27   1  [ 7] [12] [ 3] [ 3] 27
  2  +  o  .  .  13   2  [10] [ 8] [ 2] [ 2] 13
  3  .  .  *  *  53   3  [ 5] [ 4] [ 1] [ 1] 53
  4  .  .  *  *  43   4  [ 5] [ 4] [ 1] [ 1] 43
    47 39 29 21           47   39   29   21

The last combination 3C is created and also recoded:

FILE SAVE MAT P3C TO TMP / TYPE=1
...................
TRANSFORM TMP BY if(X=0)then(01)else(A) / VARS=ALL,-CASE
               A=if(X=1)then(11)else(B)
               B=if(X=2)then(13)else(C)
               C=if(X=3)then(17)else(X)
.......................................
MAT SAVE DATA TMP TO P3C
MAT RLABELS NUM(1) TO P3C

The final elimination begins at 1666 choices:

FILE EXPAND P1A2B3C
VAR z1:1,z2:1,z3:1,z4:1,z5:1,z6:1,z7:1,z8:1,z9:1 TO P1A2B3C
z1=MISSING z2=z1 z3=z2 z4=z3 z5=z4 z6=z5 z7=z6 z8=z7 z9=z8
FILE MASK P1A2B3C,z1,1,-...
...........................
VARSTAT P1A2B3C,z1,#VAL,22
VARSTAT P1A2B3C,z2,#VAL,33,34
VARSTAT P1A2B3C,z3,#VAL,51,52
VARSTAT P1A2B3C,z4,#VAL,65,204
VARSTAT P1A2B3C,z5,#VAL,39
VARSTAT P1A2B3C,z6,#VAL,44
VARSTAT P1A2B3C,z7,#VAL,55
VARSTAT P1A2B3C,z8,#VAL,26
VARSTAT P1A2B3C,z9,#VAL,1
.............................
VAR OK=1 TO P1A2B3C / SELECT=a*b*c*d*e*f*g*h*i
a=z1,0 b=z2,0 c=z3,0 d=z4,0 e=z5,1 f=z6,1 g=z7,1 h=z8,1 i=z9,1
...................
FILE MASK P1A2B3C,CASE,1,A
FILE MASK P1A2B3C,OK,1,-
............................
MAT SAVE DATA P1A2B3C TO P1A2B3C / IND=OK
MAT CLABELS NUM(1) TO P1A2B3C
MAT DIM P1A2B3C /* rowP1A2B3C=1 colP1A2B3C=16

Finally, only one choice remains:

MAT LOAD P1A2B3C 11 CUR+1
MATRIX P1A2B3C
///       1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16
4*46     26  2  3  8 39 10  1 12 11 13  7  4 17  5 44 55

By putting the column indexes (1-16) in their right places given
by diagram (b) the solution of Survo puzzle 53/2007 (c) follows.

  1A2B3C: (a)         1A2B3C: (b)                 1A2B3C: (c)
     A  B  C  D            A    B    C    D          A  B  C  D
  1  o  +  +  .  27   1  [ 7] [12] [39] [ 3] 27   1 11  8  5  3  27
  2  +  o  +  .  13   2  [10] [ 8] [26] [ 2] 13   2  6  4  1  2  13
  3  +  +  o  .  53   3  [55] [44] [17] [11] 53   3 16 15 13  9  53
  4  .  .  .  *  43   4  [ 5] [ 4] [13] [ 1] 43   4 14 12 10  7  43
    47 39 29 21           47   39   29   21         47 39 29 21

                                    Kimmo Vehkalahti (April 18, 2007)

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Solutions (May 7, 2007)

Survo puzzle 54/2007 (1)

Survo puzzle 55/2007 (240)

Survo puzzle 56/2007 (?)

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Solutions (May 28, 2007)

Survo puzzle 57/2007 (10)

Survo puzzle 58/2007 (20)

Survo puzzle 59/2007 (1400)

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Solutions (June 11, 2007)

Survo puzzle 60/2007 (10)

Survo puzzle 61/2007 (425)

Survo puzzle 62/2007 (200 after solving the hint)

This Survo puzzle has 240 different solutions, but the following hint
makes the solution unique.
Let n(k) be the number of partitions of k into prime parts.
For example n(13)=9 since 13 can be represented as a sum of prime
numbers in following ways:
1: 2+2+2+2+2+3
2: 2+2+3+3+3
3: 2+2+2+2+5
4: 2+3+3+5
5: 3+5+5
6: 2+2+2+7
7: 3+3+7
8: 2+11
9: 13

Hint:
The thirteenth root of n(83) has the continued fraction expansion
[A1,B1,C1,D1,E1,...].

Solution:
n(83) is found simply by applying the COMB operation of Survo
with a PARTS specification giving allowed parts as all possible
prime numbers:

PARTS=2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83
COMB P,CUR+1 / P=PARTITIONS,83  RESULTS=0
Partitions of 83: N[P]=12488

The thirteenth root of n(83) is

12488^(1/13)=2.0659263632126

and the start of the continued fraction expansion of this number is
obtained recursively by the following scheme

X=2.0659263632126
Y(N,X):=if(N=0)then(X)else(1/(Y(N-1,X)-A(N-1,X)))
A(N,X):=if(N=0)then(int(X))else(int(Y(N,X)))

A(0,X).=2
A(1,X).=15
A(2,X).=5
A(3,X).=1
A(4,X).=14

According to the hint we have to solve the Survo puzzle

    A   B   C   D   E
1   2  15   5   1  14  37
2   *   *   4   *   *  53
3   *   *   *   *   *  55
4  10   3   *   *   *  65
   31  42  38  31  68

and then the unique solution will be

Survo puzzle 62/2007

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Solutions (July 9, 2007)

Survo puzzle 63/2007 (10)

Survo puzzle 64/2007 (1000)

Survo puzzle 65/2007 (2050)

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Solutions (July 23, 2007)

Survo puzzle 66/2007 (120)

Survo puzzle 67/2007 (510)

Survo puzzle 68/2007 (1200)

The last problem is solved by the swapping method as follows:
The original setup is

          C1  C2  C3  C4  C5  C6 Sum  OK   D
R1        18  17  16  15  13   8  87  90  -3
R2        14  12  11   9   6   2  54  48   6
R3        10   7   5   4   3   1  30  33  -3
Sum       42  36  32  28  22  11
      OK  43  35  31  27  22  13     W=18
       D  -1   1   1   1   0  -2

and the the solution is found in six 'moves'
Move   10,12 13,14 8,9 9,10 9,11 5,6
W         14    12   8    6    4   0

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Solutions (August 20, 2007)

Survo puzzle 69/2007 (2)

Survo puzzle 70/2007 (120)

Survo puzzle 71/2007 (7300)

The problem can be solved also by the swapping method in 9 moves:
(16,17) (10,13) (5,6) (7,9) (6,7) (10,12) (10,11) (14,15) (6,8).

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Solutions (September 17, 2007)

Survo puzzle 72/2007 (0)

   A  B  C  D
1  8  *  *  *  21
2  * 10  1  *  23
3  *  *  9 11  34
  25 15 16 22

This very easy problem is solved in a following way, for example:
The only missing number in the column C is obtained by subtraction
C1=16-1-9=6:

   A  B  C  D
1  8  *  6  *  21
2  * 10  1  *  23
3  *  *  9 11  34
  25 15 16 22

Two numbers with a sum 15-10=5 in the column B are missing.
The integer 5 can be represented as a sum as 1+4 or 2+3.
The first alternative is not possible since 1 already appears as C2.
Thus the missing numbers are 2 and 3 but in which order?
If B3=3 is chosen, we have A3=34-3-9-11=11. This not allowed since
D3=11. Hence B3=2 and B1=3:

   A  B  C  D
1  8  3  6  *  21
2  * 10  1  *  23
3  *  2  9 11  34
  25 15 16 22

Now all remaining numbers are found by simple arithmetics and the
only possible solution is

   A  B  C  D
1  8  3  6  4  21
2  5 10  1  7  23
3 12  2  9 11  34
  25 15 16 22

Survo puzzle 73/2007 (500)

   A  B  C  D  E
1  *  *  *  *  *  35
2  *  *  *  *  *  27
3  *  *  *  *  *  58
  36  7 27 30 20

The task is simpler than the degree of difficulty (500) promises
because it can be solved by direct logical reasoning.
The column B must be occupied by numbers 1,2,4. Then B3=4 since only
then the sum of the row 3 can reach the value 58 which implies that
in other positions of that row the highest possible numbers 12,13,14,15
(4+12+13+14+15=58) are located:

   A   B   C   D   E
1  *   1'  *   *   *  35     numbers with a ' not necessarily
2  *   2'  *   *   *  27     in correct order row or columnwise.
3 12'  4  13' 14' 15' 58
  36   7  27  30  20

Let's study the column E. The smallest free numbers are 3 and 5.
Then the column sum 20 is possible only when E=12. All other choices
exceed the sum:

   A   B   C   D   E
1  *   1'  *   *   3' 35
2  *   2'  *   *   5' 27
3 13'  4  14' 15' 12  58
  36   7  27  30  20

Let's study the column A. The greatest free numbers are 10 and 11.
Then the column sum 36 is possible only when A3=15:

   A   B   C   D   E
1 10'  1'  *   *   3' 35
2 11'  2'  *   *   5' 27
3 15   4  13' 14' 12  58
  36   7  27  30  20

The smallest open numbers in the row 2 can be 6 and 7.
For getting a sum 27 the smallest alternatives 10,1,3 from columns A,B,E
must be selected (10+1+6+7+3=27). Other choices give a too large sum:

   A   B   C   D   E
1 11   2   *   *   5  35
2 10   1   6'  7'  3  27
3 15   4  13' 14' 12  58
  36   7  27  30  20

The only missing numbers 8 and 9 are in the row 1 and give the correct
sum 35=11+2+8+9+5.
The only possible solution will then be

   A   B   C   D   E
1 11   2   8   9   5  35
2 10   1   6   7   3  27
3 15   4  13  14  12  58
  36   7  27  30  20

Survo puzzle 74/2007 (1000)

   A  B  C  D
1  *  *  *  *  14
2  *  *  *  *  24
3  *  *  *  *  45
4  *  *  *  *  53
  20 31 40 45

The simplest solution, maybe, is obtained by using the swapping method
with the aid by the sucro /SP_SWAP (available in the newest versions
of SURVO MM). However, this procedure does not prove that there are
not other solutions.
The swapping method may be applied even without the support of /SP_SWAP.
Then one has to take the responsibility of bookkeeping.

The preliminary solution according to sizes of the products of the
marginal sums is:

          C1  C2  C3  C4 Sum  OK   D
R1         1   2   4   5  12  14  -2
R2         3   6   8  10  27  24   3
R3         7  11  13  14  45  45   0
R4         9  12  15  16  52  53  -1
Sum       20  31  40  45
      OK  20  31  40  45     W=6
       D   0   0   0   0

The total error is 6 units. The larges error (3) in in the row R2.
There is a temptation for a swap (5,8) since it eliminates that error,
but usually it is better to swap consecutive numbers so the big errors
are reduced step by step. Now small corrections are achieved by
swaps (9,10) and (5,6). The swap (9,10) gives

          C1  C2  C3  C4 Sum  OK   D
R1         1   2   4   5  12  14  -2
R2         3   6   8   9  26  24   2
R3         7  11  13  14  45  45   0
R4        10  12  15  16  53  53   0
Sum       21  31  40  44
      OK  20  31  40  45     W=6
       D   1   0   0  -1

and the the swap (5,6)

          C1  C2  C3  C4 Sum  OK   D
R1         1   2   4   6  13  14  -1
R2         3   5   8   9  25  24   1
R3         7  11  13  14  45  45   0
R4        10  12  15  16  53  53   0
Sum       21  30  40  45
      OK  20  31  40  45     W=4
       D   1  -1   0   0

It is easy to see that now the swap (2,3) solves the problem:

          C1  C2  C3  C4 Sum  OK   D
R1         1   3   4   6  14  14   0
R2         2   5   8   9  24  24   0
R3         7  11  13  14  45  45   0
R4        10  12  15  16  53  53   0
Sum       20  31  40  45
      OK  20  31  40  45     W=0
       D   0   0   0   0

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Solutions (October 15, 2007)

Survo puzzle 75/2007 (60)

Survo puzzle 76/2007 (680)

Survo puzzle 77/2007 (?)

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Solutions (October 29, 2007)

Survo puzzle 78/2007 (55)

Survo puzzle 79/2007 (175)

Survo puzzle 80/2007 (17500)

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Solutions (November 12, 2007)

Survo puzzle 81/2007 (50)

Survo puzzle 82/2007 (400)

Survo puzzle 83/2007 (20000)

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Solutions (November 26, 2007)

Survo puzzle 84/2007 (15)

Survo puzzle 85/2007 (25)

Survo puzzle 86/2007 (50)

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Solutions (December 10, 2007)

Survo puzzle 87/2007 (120)

Survo puzzle 88/2007 (400)

Survo puzzle 89/2007 (1500)

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Solutions (December 24, 2007)

Survo puzzle 90/2007 (400)

Survo puzzle 91/2007 (?)

Survo puzzle 92/2007 (9800)

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Solutions (January 7, 2008)

Survo puzzle 93/2007 (790)

Survo puzzle 94/2007 (90)

Survo puzzle 95/2007 (790)

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Solutions (January 28, 2008)

Survo puzzle 96/2008 (250)

Survo puzzle 97/2008 (250)

Survo puzzle 98/2008 (1800)

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Solutions (February 11, 2008)

Survo puzzle 99/2008 (220)

Survo puzzle 100/2008 (630)

Survo puzzle 101/2008 (2000)

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Solutions (February 25, 2008)

Survo puzzle 102/2008 (130)

Survo puzzle 103/2008 (150)

Survo puzzle 104/2008 (1900)

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Solutions (March 10, 2008)

Survo puzzle 105/2008 (160)

Survo puzzle 106/2008 (320)

Survo puzzle 107/2008 (500)

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Solutions (March 24, 2008)

Survo puzzle 108/2008 (30)

Survo puzzle 109/2008 (220)

Survo puzzle 110/2008 (430)

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Solutions (April 7, 2008)

Survo puzzle 111/2008 (45)

Survo puzzle 112/2008 (160)

Survo puzzle 113/2008 (850)

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Solutions (May 5, 2008)

Survo puzzle 114/2008

Survo puzzle 115/2008

Survo puzzle 116/2008

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Solutions (June 2, 2008)

Survo puzzle 117/2008

Survo puzzle 118/2008

Survo puzzle 119/2008

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Solutions (June 30, 2008)

Survo puzzle 120/2008

Survo puzzle 121/2008

Survo puzzle 122/2008

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Solutions (July 28, 2008)

Survo puzzle 123/2008

Survo puzzle 124/2008

Survo puzzle 125/2008

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Solutions (August 25, 2008)

Survo puzzle 126/2008

Survo puzzle 127/2008

Survo puzzle 128/2008

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Solutions (September 22, 2008)

Survo puzzle 129/2008

Survo puzzle 130/2008

Survo puzzle 131/2008

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Solutions (October 20, 2008)

Survo puzzle 132/2008

Survo puzzle 133/2008

Survo puzzle 134/2008

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Solutions (November 17, 2008)

Survo puzzle 135/2008

Survo puzzle 136/2008

Survo puzzle 137/2008

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Solutions (December 15, 2008)

Survo puzzle 138/2008

Survo puzzle 139/2008

Survo puzzle 140/2008

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Solutions (January 12, 2009)

Survo puzzle 141/2008

Survo puzzle 142/2008

Survo puzzle 143/2008

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Solutions (February 9, 2009)

Survo puzzle 144/2009

Survo puzzle 145/2009

Survo puzzle 146/2009

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Solutions (March 9, 2009)

Survo puzzle 147/2009

Survo puzzle 148/2009

Survo puzzle 149/2009

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Solutions (April 6, 2009)

Survo puzzle 150/2009

Survo puzzle 151/2009

Survo puzzle 152/2009