14.11.2007, 12:19

OK Eisbär (and hopefully others),

Here we go:

(Notation: Row 1, Column 1 is R1C1)

In this case the lower grid gives the best opportunities to start.

There is one X in this grid. It must belong to SEMAX. So the S and E above the X, and the M right of the X belong to the same pentomino. The only missing letter is an A. If you should take the A right of the M, on the bottom row there would be (starting in R5C1) A, J, A, A together. And there is no name with those letters in it. So take the A down the X to complete SEMAX. The first pentomino is finished.

There are two Bs in the grid. The B in R4C12 belongs to CHALB. So the L, A, C above the B belong to the same pentomino. If you should complete the name with the H right of the A, you would get the same Pentomino as SEMAX. That is not allowed, so take the H left of the B. Second pentomino is finished.

R in R5C12 belongs to the two Es left of it. EER can only belong to PETER, T and P above the E. Third name, third Pentomino.

Now the O in R3C11 goes together with the H above it. Can only be HOPET, with the T and E left of the H, and the P beneath the E. Fourth name, fourth pentomino.

Almost another name/pentomino is formed now: The letters KAUS in R1C8-11 are the main part of KRAUS. Connection can only be made with the R left of the K, because of the CHALB-pentomino.

The letter F appears three times in the lower grid. They belong to ZAFER, FRANK and MAFER. The F in R2C2 cant reach a M or K and belongs to ZAFER. Connect with the Z in R1C1 via the A above this F. The next F (in R1C5) cant reach an M, so must belong to FRANK. Connect F with A next to it, N beneath A and K right of the N. Two possibilities left for the remaining R.

The B in R4C8 belongs to BAARS and uses two As. The I in R3C7 belongs to MARIA and uses also two As. Therefore the A in MAFER must be the A in R5C5. MAFER now consists of F in R3C6, the E left of it, the R and M beneath the E and the A left of the M.

BAARS and MARIA use together 10 letters. So the R in R2C8 does not belong to MARIA or BAARS and therefore it belongs to FRANK, and another Pentomino is finished.

Now BAARS and MARIA can be finished. It is only in one way possible to avoid a shape to be used twice.

TANJA, starting with the J in R5C2 can also be finished: J, A right of it, A above the A, T right of the second A and N in R3C3. If the T in R3C4 is used, the MARIA-shape appears twice.

Finally, IRENE and ZAFER can be formed with the two remaining shapes.

Now the lower grid is solved, 12 names are used and the upper grid is next. It goes in a similar way, but is a bit harder. Starting point is ROLLO in the left side of the grid.

'Viel Spass' in solving!

If other people enjoy this puzzle as well, I can make another one, but I have to think about the subject.

Regards,

Richard

Here we go:

(Notation: Row 1, Column 1 is R1C1)

In this case the lower grid gives the best opportunities to start.

There is one X in this grid. It must belong to SEMAX. So the S and E above the X, and the M right of the X belong to the same pentomino. The only missing letter is an A. If you should take the A right of the M, on the bottom row there would be (starting in R5C1) A, J, A, A together. And there is no name with those letters in it. So take the A down the X to complete SEMAX. The first pentomino is finished.

There are two Bs in the grid. The B in R4C12 belongs to CHALB. So the L, A, C above the B belong to the same pentomino. If you should complete the name with the H right of the A, you would get the same Pentomino as SEMAX. That is not allowed, so take the H left of the B. Second pentomino is finished.

R in R5C12 belongs to the two Es left of it. EER can only belong to PETER, T and P above the E. Third name, third Pentomino.

Now the O in R3C11 goes together with the H above it. Can only be HOPET, with the T and E left of the H, and the P beneath the E. Fourth name, fourth pentomino.

Almost another name/pentomino is formed now: The letters KAUS in R1C8-11 are the main part of KRAUS. Connection can only be made with the R left of the K, because of the CHALB-pentomino.

The letter F appears three times in the lower grid. They belong to ZAFER, FRANK and MAFER. The F in R2C2 cant reach a M or K and belongs to ZAFER. Connect with the Z in R1C1 via the A above this F. The next F (in R1C5) cant reach an M, so must belong to FRANK. Connect F with A next to it, N beneath A and K right of the N. Two possibilities left for the remaining R.

The B in R4C8 belongs to BAARS and uses two As. The I in R3C7 belongs to MARIA and uses also two As. Therefore the A in MAFER must be the A in R5C5. MAFER now consists of F in R3C6, the E left of it, the R and M beneath the E and the A left of the M.

BAARS and MARIA use together 10 letters. So the R in R2C8 does not belong to MARIA or BAARS and therefore it belongs to FRANK, and another Pentomino is finished.

Now BAARS and MARIA can be finished. It is only in one way possible to avoid a shape to be used twice.

TANJA, starting with the J in R5C2 can also be finished: J, A right of it, A above the A, T right of the second A and N in R3C3. If the T in R3C4 is used, the MARIA-shape appears twice.

Finally, IRENE and ZAFER can be formed with the two remaining shapes.

Now the lower grid is solved, 12 names are used and the upper grid is next. It goes in a similar way, but is a bit harder. Starting point is ROLLO in the left side of the grid.

'Viel Spass' in solving!

If other people enjoy this puzzle as well, I can make another one, but I have to think about the subject.

Regards,

Richard