Decode the following message. Each letter could be thought of as an integer on
the number line ( A=1,b=2 etc). If you arrange the letters so that all the
inequalities are true, the message will be clear.
S < R F < O R < L
T > S N > U C < H
L < U T < A N < C
H > L O < H U > A
Y < F Y > A R > O
I know that the answer is "Stay for Lunch" however I can't figure out how they
derived that answer. This problem has been frustrating me for hours and
hours!!! PLEASE HELP
First reverse all the "greater than" inequalities to their equivalent
"less than" inequalities:
S < R
T > S change to S < T
L < U
H > L change to L < H
Y < F
F < O
N > U change to U < N
T < A
O < H
Y > A change to A < Y
R < L
C < H
N < C
U > A change to A < U
R > O change to O < R
Relist them:
S < R
S < T
L < U
L < H
Y < F
F < O
U < N
T < A
O < H
A < Y
R < L
C < H
N < C
A < U
O < R
Notice that S does not appear on the right of any " < ".
Therefore none of the letters are less than S. Therefore
S is the very first letter of the message.
Notice that H does not appear on the left of any " < ".
Therefore none of the letters are greater than H. Therefore
H is the very last letter of the message.
So the message so far is
S < _ < _ < _ < _ < _ < _ < _ < _ < _ < _ < _ < _ < _ < _ < H
we have to fill in the blanks to form this "chain" of
letters from S to H.
We must find any "link" in the "chain". That is, we must
find one of the inequalities in the list such that we are
certain that the letter on the left of < and the letter
on the right of the < come next to each other in the
message.
We cannot be sure whether S and R or S and T are next to
each other in the message because both S < R and S < T appear
in the list.
Also, by the same token of reasoning it's the same with L because
both L < U and L < H appear, and also with A because A < Y and A < U
both appear. Also with R because S < R and O < R appear. Also
with U because L < U and A < U, and with H, which appears 3 times on
the right.
So the first inequality on the list that appears only once on
the left and also only once on the right is Y < F, so we write
that down and call it "partial chain 1", and mark Y < F off the
list:
partial chain 1: Y < F
Now since F appears only once on the left of a < as F < O, we can
annex "< O" on the right of the preceding partial chain and now
we have this partial chain, and mark F < O off the list:
partial chain 1: Y < F < O
But we can go no further on the right because O appears twice
on the left of an inequality.
So we see if we can annex anything on the left of Y. We see that
Y appears only once on the right of < as A < Y. So we can annex
"A <" on the left side of our partial chain, and mark A < Y off the
list:
partial chain 1: A < Y < F < O
We can also mark A < U off the list, for we have placed A in a partial
chain.
Next we see if we can annex anything on the left of A. We see that
A appears only once on the right of < as T < A. So we can annex
"T <" on the left side of our partial chain 1, and mark T < A off the
list:
partial chain 1: T < A < Y < F < O
Next we see if we can annex anything on the left of T. We see that
T appears only once on the right of < as S < T. So we can annex
"S <" on the left side of our partial chain 1, so mark S < T off the
list. We can also mark off S < R for we are done with S:
partial chain 1: S < T < A < Y < F < O
Since S is the very first letter, we can't extend the chain
any further left. And we have already shown that we cannot extend
it on the right (without guessing). So we must make a new partial
chain. The next inequality we haven't used whose left side appears
only once on the left and whose right side appears only once on the
right is U < N. So we must
start a new partial chain:
partial chain 2: U < N
So we mark U < N off the list. Next we see if we can annex anything
on the right of N. We see that N appears only once on the left of
< as N < C. So we can annex "< C" on the right side of our partial
chain 2, and mark N < C off the list:
partial chain 2: U < N < C
Next we see if we can annex anything on the right of C. We see that
C appears only once on the left of < as C < H. So we can annex
"< H" on the right side of our partial chain 2, and mark C < H off
the list:
partial chain 2: U < N < C < H
Since H is the last letter in the message, we can go no further on
the right. We can also mark L < H off the list, since we have placed H
in a partial list.
Next we see if we can annex anything on the left of U. Since we have
marked off A < U, U appears on the right only as L < U. So we can annex
"< H" on the right side of our partial chain 2, and mark L < U off the
list.
partial chain 2: L < U < N < C < H
We also mark off L < H from the list because we have placed "L"
in a partial chain. We can also mark O < H from the list since
we have placed H in a partial chain.
Next we see if we can annex anything on the left of L. Since L
appears on the right only as R < L. So we can annex
"R <" on the left side of our partial chain 2, and mark R < L off the
list.
partial chain 2: R < L < U < N < C < H
Next we see if we can annex anything on the left of R. Since S < R
has been scratched off, R appears on the right only as 0 < R. So we can annex
"O <" on the left side of our partial chain 2.
partial chain 2: O < R < L < U < N < C < H
But, hey, O is at the end of partial chain 1, so now we can put
partial chain 1 left of partial chain 2, and we get
S < T < A < Y < F < O < R < L < U < N < C < H
Edwin
