# proofs without words

Published articles for proofs without words.

This is one page of public article previews, not the complete archive. Follow Next page to continue. Summaries are not the original full articles.

## Tiling a Chessboard with Dominoes (Opposite Colors Removed)

DevFeed: [Tiling a Chessboard with Dominoes (Opposite Colors Removed)](<https://devfeed.tech/articles/tiling-a-chessboard-with-dominoes-opposite-colors-removed-40248.md>)

Original publisher: [Read original article](<https://www.jeremykun.com/2011/11/18/tiling-a-chessboard-2/>)

Published: 2011-11-18T10:32:54Z

Content type: tutorial

Language: en

Sources: [Jeremy Kun](<https://devfeed.tech/sources/jeremy-kun.md>)

Topics: [Algorithm](<https://devfeed.tech/topics/algorithm.md>)

Tags: [algorithm](<https://devfeed.tech/tags/algorithm.md>), [analysis](<https://devfeed.tech/tags/analysis.md>), [chessboard](<https://devfeed.tech/tags/chessboard.md>), [color](<https://devfeed.tech/tags/color.md>), [dominoes](<https://devfeed.tech/tags/dominoes.md>), [proofs-without-words](<https://devfeed.tech/tags/proofs-without-words.md>), [tiling](<https://devfeed.tech/tags/tiling.md>)

### AI overview

A constructive solution shows that a chessboard with two opposite-colored squares removed can be tiled with 2-by-1 dominoes. The article explains the placement pattern and notes that the corresponding algorithm runs in linear time.

### Source excerpt

This is a natural follow-up to our first gallery entry on the impossibility of tiling certain chessboards with dominoes. Problem: Suppose we remove two squares from a chessboard which have opposite color. Is it possible to tile the remaining squares with 2-by-1 dominoes? Solution: Notice that if we remove two squares of opposite color, then there is only one way to place dominoes on the remaining squares according to this scheme (one cannot tile a domino across the "walls").

## N Choose 2 is the Sum of the First N-1 Integers

DevFeed: [N Choose 2 is the Sum of the First N-1 Integers](<https://devfeed.tech/articles/n-choose-2-is-the-sum-of-the-first-n-1-integers-40243.md>)

Original publisher: [Read original article](<https://www.jeremykun.com/2011/10/02/n-choose-2/>)

Published: 2011-10-02T16:16:47Z

Content type: tutorial

Language: en

Sources: [Jeremy Kun](<https://devfeed.tech/sources/jeremy-kun.md>)

Topics: [Mathematics](<https://devfeed.tech/topics/mathematics.md>)

Tags: [arithmetic](<https://devfeed.tech/tags/arithmetic.md>), [bijections](<https://devfeed.tech/tags/bijections.md>), [combinatorics](<https://devfeed.tech/tags/combinatorics.md>), [mathematics](<https://devfeed.tech/tags/mathematics.md>), [proofs-without-words](<https://devfeed.tech/tags/proofs-without-words.md>)

### AI overview

A combinatorial bijection shows that the binomial coefficient n choose 2 equals the sum of the first n−1 integers. The article explains how yellow dots correspond uniquely to pairs of dots in the bottom row, then briefly connects bijections to isomorphism and classification in mathematics.

### Source excerpt

Problem: Determine an arithmetic expression for $ \binom{n}{2}$. Solution: The following picture describes a bijection between the set of yellow dots and the set of pairs of purple dots: In particular, selecting any yellow dots and travelling downward along diagonals gives a unique pair of blue dots. Conversely, picking any pair of blue dots gives a unique yellow dot which is the meeting point (the "peak") of the inward diagonals. If we say the bottom row has $ n$ elements, then the number of yellow dots is clearly $ 1 + 2 + \dots + (n-1)$, and the number of pairs in the last row is just $ \binom{n}{2}$.