In this grid horizontal and vertical words are symmetrical about the upper left to lower right diagonal.
T R Y
R Y E
Y E S
This grid is to be reformed by replacing all letters in 6 steps having regard to symmetry, changing one letter at a time and using only English words.
While my solution reflects my title at which you might not arrive, can you at least publish a new grid under my rules?
Asides:
1. Show us any 3 x 3 grids which you have composed from a significantly different initial grid and their development.
2. Beginning with a 4 x 4 grid is it possible to arrive at a new grid, same rules, in 10 steps?
NOTE:
Because of the nature of the language - "English words" - is a difficult phrase to define. If a word seems to need justification then please do so; challenges are otherwise likely.
(In reply to
re: so many ways by brianjn)
Actually, was done via computer program:
DECLARE SUB changeIt ()
DECLARE FUNCTION isWord! (w$)
DIM SHARED hist$(10, 3), w$(3), did(3, 3), howMany, ct
CLS
w$(1) = "try"
w$(2) = "rye"
w$(3) = "yes"
OPEN "wordsq.txt" FOR OUTPUT AS #4
changeIt
PRINT ct
CLOSE
SUB changeIt
FOR row = 1 TO 3
FOR col = row TO 3
IF did(row, col) = 0 THEN
FOR subs = 1 TO 26
letr$ = MID$("abcdefghijklmnopqrstuvwxyz", subs, 1)
h$ = MID$(w$(row), col, 1)
MID$(w$(row), col, 1) = letr$
IF row <> col THEN
MID$(w$(col), row, 1) = letr$
END IF
good = 1
IF letr$ = h$ THEN good = 0
IF isWord(w$(row)) = 0 THEN
good = 0
ELSE
IF row <> col THEN
IF isWord(w$(col)) = 0 THEN good = 0
END IF
END IF
IF good THEN
did(row, col) = 1
howMany = howMany + 1
hist$(howMany, 1) = w$(1)
hist$(howMany, 2) = w$(2)
hist$(howMany, 3) = w$(3)
IF howMany = 6 THEN
FOR i = 1 TO 6
PRINT hist$(i, 1); " ";
NEXT
PRINT
FOR i = 1 TO 6
PRINT hist$(i, 2); " ";
NEXT
PRINT
FOR i = 1 TO 6
PRINT hist$(i, 3); " ";
NEXT
PRINT
PRINT
FOR i = 1 TO 6
PRINT #4, hist$(i, 1); " ";
NEXT
PRINT #4,
FOR i = 1 TO 6
PRINT #4, hist$(i, 2); " ";
NEXT
PRINT #4,
FOR i = 1 TO 6
PRINT #4, hist$(i, 3); " ";
NEXT
PRINT #4,
PRINT #4,
ct = ct + 1
ELSE
changeIt
END IF
howMany = howMany - 1
did(row, col) = 0
END IF
MID$(w$(row), col, 1) = h$
IF row <> col THEN
MID$(w$(col), row, 1) = h$
END IF
NEXT subs
END IF
NEXT
NEXT
END SUB
FUNCTION isWord (w$)
n = LEN(w$)
IF w$ = "i" THEN isWord = 1: EXIT FUNCTION
w1$ = SPACE$(n)
OPEN "\words\comword3.bin" FOR BINARY AS #2
l = LOF(2) / n
low = 1: high = l
DO
mid = INT((low + high) / 2)
GET #2, (mid - 1) * n + 1, w1$
IF w1$ = w$ THEN isWord = 1: CLOSE 2: EXIT FUNCTION
IF w1$ < w$ THEN low = mid + 1: ELSE high = mid - 1
LOOP UNTIL low > high
isWord = 0
CLOSE 2
END FUNCTION
and was done yesterday during daylight hours and did not detract from sleep.
|
Posted by Charlie
on 2007-04-26 09:53:32 |