I have an algorithm that has the following cost:
C(Alg) <= t * Z
I've found on the web an exercise that state the following :
if t>=x, I can say that C(Alg) <= x * Z <= t * Z
but it looks strange.. do you agree?
Asked
Active
Viewed 47 times
0
tshepang
- 12,111
- 21
- 91
- 136
user3158123
- 63
- 4
-
2This question appears to be off-topic because it is about [math.se]. – Bernhard Barker Jan 03 '14 at 17:35
-
Consider x=0 and t=1. One doesn't have to think very much to see that what you're asking is false. – Paul Hankin Jan 03 '14 at 18:02
1 Answers
0
This is in general not true. Think about the case where $Z$ is negative, then this does not hold true:
$$\text{if} x \leq Z, \text{then} xZ \leq tZ$$
tonga
- 11,749
- 25
- 75
- 96
-
-
Even if Z is always positive, the conclusion that `C(Alg)<=x*Z<=t*Z` under the condition that `x<=t` is incorrect because `xZ` is a tighter upper bound. – tonga Jan 03 '14 at 17:46