First off, read what george wrote - it's important to understand the
context in which an access-list is being used. What works for one
context may not work for another, with an excellent example being
packet filtering vs. route filtering.

When you understood that, you need to look at the bit differences (a
student in the last class told me a fancy name for it, but Mad Cow
disease made me forget).

Let's take your example:

> 192.168.1.15   1100 0000. 1010 1000. 0000 0001. 0000 1111
> 192.168.1.29   1100 0000. 1010 1000. 0000 0001. 0001 1101

You have two addresses that you want to match with a single ACL line.
These two have TWO bit differences. Bits 1 and 4 in the last octet:

15 = 00001111
29 = 00011101

Looking from right to left, bits are numbered from 0 to 7 in the
octet. These are also powers of 2:

Bit 0 = 2^0 (1)
Bit 1 = 2^1 (2)
Bit 2 = 2^2 (4)
Bit 3 = 2^3 (8)
Bit 4 = 2^4 (16)
Bit 5 = 2^5 (32)
Bit 6 = 2^6 (64)
Bit 7 = 2^7 (128)

When you have N bit differences, the total number of possible
combinations is 2^N. In this case, we have two bit differences, which
means that maximum number of combinations is 2^2=4. That means that
the only way to write ACL in a single line and encompass these two
addresses is to include AT LEAST two others.

Given this logic, without calculations, you can only determine if it's
POSSIBLE to include several addresses with a single line. Here's a
quick cheat:

1. If number of addresses to match is a power of two, it COULD be possible
2. If the bit difference is such that 2^(bit difference) matches the
total number of networks it COULD be possible
3. If the difference bits are on the same positions and #1 and #2 are
true, it IS possible

--
Marko Milivojevic - CCIE #18427 (SP R&S)
Senior CCIE Instructor - IPexpert

On Wed, Jan 11, 2012 at 07:17, kyujin Choi <[email protected]> wrote:
> only one wild card statement vs two wild card statement
>
> I am practicing wild card. I need to mention what I understand first in
> order to get better answers.
>
> I am missing an ability whether I can express network range by using 1 wild
> card statement, or not. (I put examples below)
>
> ---------------------------------------------------------------
> This is what I understand for wild card for 192.168.1.15 - 192.168.1.29
>
> 192.168.1.15   1100 0000. 1010 1000. 0000 0001. 0000 1111
> 192.168.1.29   1100 0000. 1010 1000. 0000 0001. 0001 1101
>
> (same part)    1100 0000. 1010 1000. 0000 0001. 000x xxxx (192.168.1.0)
> (wild card)    0000 0000 .0000 0000 .0000 0000 .31 is wild card.
>
> 192.168.1.0 0.0.0.31
>
> I understand this above.
>
>
> -----------------------
>
> 10.1.2.0/24 - 10.1.3.0/24
>
> (network by using wild card)
> 10.1.2.0 0.0.1.255
>
> I understand this above, too.
>
>
> -----------------------------
>
> Question) I do understnad this below, but is there any easy way to remember
> whether I can express range network through
> only one wild card statement or not.
>
> 10.1.1.0/24 - 10.1.2.0/24
> (what book said)
> 10.1.1.0 0.0.0.255
> 10.1.2.0 0.0.0.255
>
>
> (what I thought)
> 10.1.0.0 0.0.3.255   (but this statement covers more than that;  10.1.0.0 -
> 10.1.3.255)
>
> ----------------------
> Another example
>
> 192.168.32.0/24 - 192.168.40.0/24
>
> (what book said)
> 192.168.32.0 0.0.7.255
> 192.168.40.0 0.0.0.255
>
> (what I thought)
> 192.168.32.0 0.0.15.255  (but this statment covers more that that again;
> 192.168.32.0 - 192.168.47.255)
>
>
> Consequently, I am missing an ability whether I can express network range
> by using 1 wild card statement, or not.
> Is there any easy way I can figure it out quickly whether I need more than
> 1 wild card statement?
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>
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>
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_______________________________________________
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