# Difference between revisions of "TADM2E 3.12"

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− | Let's put into the black box whole set <math>S=\{x_i\}_{i=1}^n</math>. If <math>bb(S)</math> is True, then such subset | + | Let's put into the black box whole set <math>S=\{x_i\}_{i=1}^n</math>. If <math>bb(S)</math> is True, then such a subset exists and we can go on: |

# R:=S | # R:=S | ||

# for i:=1 to n do | # for i:=1 to n do | ||

## If <math>bb(R/\{x_i\})</math> is True then <math>R:=R/\{x_i\}</math> | ## If <math>bb(R/\{x_i\})</math> is True then <math>R:=R/\{x_i\}</math> | ||

When this iteration is finished R will be subset of S that adds up to k. | When this iteration is finished R will be subset of S that adds up to k. | ||

+ | |||

+ | Above solution works even when there are multiple subsets that add up to k. |

## Latest revision as of 08:59, 26 December 2019

Let's put into the black box whole set $ S=\{x_i\}_{i=1}^n $. If $ bb(S) $ is True, then such a subset exists and we can go on:

- R:=S
- for i:=1 to n do
- If $ bb(R/\{x_i\}) $ is True then $ R:=R/\{x_i\} $

When this iteration is finished R will be subset of S that adds up to k.

Above solution works even when there are multiple subsets that add up to k.