HMMT 十一月 2009 · 冲刺赛 · 第 18 题
HMMT November 2009 — Guts Round — Problem 18
题目详情
英文原题
- [ 9 ] Let f be a function that takes in a triple of integers and outputs a real number. Suppose that fsatisfies the equationsf ( a + 1 , b, c ) + f ( a − 1 , b, c )
f ( a, b, c ) =
f ( a, b + 1 , c ) + f ( a, b − 1 , c )2
f ( a, b, c ) =
f ( a, b, c + 1) + f ( a, b, c − 1)2
f ( a, b, c ) =
for all integers a, b, c . What is the minimum number of triples at which we need to evaluate f in order 2
to know its value everywhere?
2 ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUNDnd
解析
英文解析
- [ 9 ] Let f be a function that takes in a triple of integers and outputs a real number. Suppose that fsatisfies the equationsf ( a + 1 , b, c ) + f ( a − 1 , b, c )
f ( a, b, c ) =
f ( a, b + 1 , c ) + f ( a, b − 1 , c )2
f ( a, b, c ) =
f ( a, b, c + 1) + f ( a, b, c − 1)2
f ( a, b, c ) =
for all integers a, b, c . What is the minimum number of triples at which we need to evaluate f in order 2
to know its value everywhere?
Answer: 8 Note that if we have the value of f at the 8 points:
(0 , 0 , 0) , (1 , 0 , 0) , (0 , 1 , 0) , (0 , 0 , 1) , (0 , 1 , 1) , (1 , 0 , 1) , (1 , 1 , 0) , (1 , 1 , 1), we can calculate the value for anytriple of points because we have that f ( a + 1 , b, c ) − ( a, b, c ) constant for any a , if b and c are fixed
(and similarly for the other coordinates). To see why we cannot do this with less points, notice thatwe need to determine what the value of these 8 points anyways, and there is no ”more efficient” wayto determine them all in fewer evaluations.
2 ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 7 NOVEMBER 2009 — GUTS ROUNDnd